Ch 2.1 D=R.T
distance
speed =
time
v ∆x
v=
( x → l , d , s)
∆t
Ex 2.1 p23) A student drives from home to school at an average
speed of 25.3 km/hr. If it takes her 4.72 hr to get there, how far did
she travel?
To convert km/hr to m/s divide by 3.6
Ex 2.2) The moon’s nearly circular orbit around the earth has
a radius of 3.84x108 m. If it takes 27.3 days to complete 1 revolution,
what is its average orbital speed?
Ex 2.4) A 600 km race is won by a team of 2 drivers, each had the
wheel for half the distance. Driver 1 averaged 60 km/hr, driver 2
averaged 20 km/hr. What is their overall average speed?
HW #18) To test a small rocket engine, it’s fired straight up 100 m
in 5 sec, then it falls back to the ground at an average in speed
of 10 m/s. How long did the whole trip take? What was the net
average speed?
Ch2 HW#1 pg 49+
1,3,4,5,9,18
Ch2 HW#1 p49+ 1,3,4,5,9,18
1. In the next 3s, the Earth will travel roughly 885,000m
along its orbit around the Sun. Compute its average
orbital speed.
3. Standing on the roof of a building a kid drops a plastic
bag filled with water at a height of 100 m, and 4.5 s
later it strikes the ground. Determine the bag’s
average speed.
4. Hair grows at an average rate of 3 x 10-9 m/s.
How long will it take to grow a 10 cm strand?
5. Given that a glacier creeps along at an average speed
of 1x10-6 m/s, how long will it take to advance 1.0km?
9. If the odometer in a car at the beginning of a trip read
12723.10km and 2.00 h later it read 12973.10km,
what was the average speed during the journey?
Up
Down
18. To test a small rocket motor it’s fired up a long
vertical tube. It rises to a height of 100 m in 5.0 s and
then falls back to the ground at an average speed of
10.0 m/s. How long did the whole trip take, and what
was the net average speed?
Total time = tup + tdown
100m
= 10 s
tup=5 s t down =
10m / s
Vave=
Ch2.2 Graphing Speed
Ex 1) A bee travels at a constant speed.
After 1 sec, it’s gone 10 m
2 sec, it’s gone 20 m
3 sec, it’s gone 30 m
What is it’s constant speed?
40
30
Dist
(m)
20
10
1
2
3
time (sec)
4
5
Ch2.2 Graphing Speed
Ex 1) A bee travels at a constant speed.
After 1 sec, it’s gone 10 m
2 sec, it’s gone 20 m
3 sec, it’s gone 30 m
What is it’s constant speed?
40
30
Dist
(m)
20
10
1
The slope of a distance-time graph gives a speed.
2
3
time (sec)
4
5
Ch2.2 Graphing Speed
Ex 2) If the bee continues with this same
constant speed, how far will it travel in 5 sec?
40
30
Speed
(m/s)
20
10
1
2
3
time (sec)
4
5
Ch2.2 Graphing Speed
Ex 2) If the bee continues with this same
constant speed, how far will it travel in 5 sec?
40
30
Speed
(m/s)
20
10
1
The area under the curve of the velocity-time graph
gives the distance traveled.
2
3
time (sec)
4
5
Ex 3) If the bee’s speed is changing
constantly as graphed here, what is
its
instantaneous speed
at 5 sec?
at 7 sec?
Dist
(m)
12
10
8
6
4
2
1
2
3 4 5 6
time (sec)
7
8
Ex 3) If the bee’s speed is changing
constantly as graphed here, what is
its instantaneous speed
at 5 sec?
at 7 sec?
Dist
(m)
12
10
8
6
4
2
1
2
3 4 5 6
time (sec)
7
8
To find the instantaneous speed, take the slope of the line tangent to that point.
Ex 4) What was the bee’s
average velocity over:
the 1st sec?
the 1st 3 sec?
over 8 sec?
What’s the bee doing at 2 secs?
Dist
(m)
12
10
8
6
4
2
1
2
3 4 5 6
time (sec)
7
8
Ex 4) What was the bee’s
average velocity over:
the 1st sec?
the 1st 3 sec?
over 8 sec?
What’s the bee doing at 2 secs?
Dist
(m)
12
10
8
6
4
2
1
2
To find the average velocity:
1. Find the total distance and the total time then divide.
2. Find the slope of the line that connects the endpoints.
3 4 5 6
time (sec)
7
8
Ch2 HW#2 p50 20,21,23,24,25,27,32,33
23. Figure P23 is a plot of the speed of a
cat versus time. How far did the cat
travel during the third second of its
journey? What were its maximum and speed
(m/s)
minimum speeds?
When, if ever, was its speed a nonzero
constant?
6
5
4
3
2
1
1
2
3 4 5
time (sec)
6
7
7
6
27. Figure P27 shows the distance traveled
versus time for a toy car. What was the
toy car’s average speed during the time
interval from 2.0 s to 8.0 s?
5
dist
4
(m)
3
2
1
1
2
3 4 5
time (sec)
6
7
8
Ch2 HW#2 p50 20,21,23,24,25,27,32,33
23. Figure P23 is a plot of the speed of a
cat versus time. How far did the cat
travel during the third second of its
journey? What were its maximum and speed
(m/s)
minimum speeds?
When, if ever, was its speed a nonzero
constant?
Max: 4-5sec
Min: 0,7sec
5
4
3
2
1
1
2
3 4 5
time (sec)
6
7
7
Const @: 1.5-2, 2-3, 4-5,
27. Figure P27 shows the distance traveled
versus time for a toy car. What was the
toy car’s average speed during the time
interval from 2.0 s to 8.0 s?
6
6
5
dist
4
(m)
3
2
1
1
2
3 4 5
time (sec)
6
7
8
Ch2 HW#2 p50+ 20,21,23,24,25,27,32,33
20. You know how you close your eyes when you
sneeze? Suppose you are driving along at a constant
96.5 km/h (i.e. 60 mi/h) and you experience a 1.00 s
long, eyes-closed, giant sneeze. How many meters
does the car travel while you are out of control?
21. Light travels in a vacuum at a fixed speed of roughly
2.998 x 108 m/s and its speed in air is only negligibly
slower.
a. How long does it take light to traverse 1ft (~.30m )?
b. That means that when you are looking at
something 1000 m away, you are seeing it as it was
____second(s) back in time.
24. The Earth rotates once around its spin axis at 23 h
56 min and its equatorial diameter is 1.276 x 107 m
(i.e. 7927 mi). At what speed would you be traveling
with respect to the stars on the Equator?
25. Use Fig 2.5 (page 26) to calculate the distance
traveled by the bee (whose speed-time graph is
plotted) in the time interval from 1.33 s to 2.83 s.
30
v 20
10
1 2 3
t
32. Figure P23 is a plot of the speed of a cat versus time. 6
Approximately, what was the instantaneous speed
5
at each of the following times:
speed 4
0s
(m/s) 3
1.0 s
2
2.0 s
1
4.5 s
6.0 s
1 2 3 4 5 6 7
7.0 s?
time (sec)
During what time intervals was the speed increasing?
When was its speed decreasing?
700
33. While driving along a winding mountain road
600
a passenger makes a plot of the tripmeter’s
500
readings against time.
Placing a dot on the distance versus time curve dist 400
every 10.0 s for 5 min, he gets a straight line
(m) 300
passing through the x = 0, t = 0 origin
200
having a slope of 16.0 m/s.
100
What is the instantaneous speed of the car at
10 20 30 40 50 60 70 80
45 s into the exercise?
time (sec)
How far does the car travel in the time between
t = 86 s and t = 186 s?
Ch 2.3 – Velocity
Velocity - has speed (magnitude) and direction
scalar
vector
- use arrows to represent vectors
Ch 2.3 – Velocity
Velocity - has speed (magnitude) and direction
scalar
vector
- use arrows to represent vectors
Displacement – straight line drawn between
initial position (xi) and final position (xf)
Has both magnitude and direction,
so displacement is a vector quantity.
Don’t use distance and displacement interchangeably.
Your car odometer measures distance.
“As the crow flies” measures displacement.
xf
xi
Ch 2.3 – Velocity
Velocity - has speed (magnitude) and direction
scalar
vector
- use arrows to represent vectors
Displacement – straight line drawn between
initial position (xi) and final position (xf)
Has both magnitude and direction,
so displacement is a vector quantity.
Don’t use distance and displacement interchangeably.
Your car odometer measures distance.
“As the crow flies” measures displacement.
xf
xi
Ch 2.3 – Velocity
Velocity - has speed (magnitude) and direction
scalar
vector
- use arrows to represent vectors
xf
Displacement – straight line drawn between
initial position (xi) and final position (xf)
Has both magnitude and direction,
so displacement is a vector quantity.
Don’t use distance and displacement interchangeably.
Your car odometer measures distance.
“As the crow flies” measures displacement.
HW pg 51 #36) A mouse runs straight north 1.414 m,
stops, turn right at 90 and runs another 1.414 m.
Through what distance was it displaced?
xi
1.414 m
1.414 m
Ch 2.3 – Velocity
Velocity - has speed (magnitude) and direction
scalar
vector
- use arrows to represent vectors
xf
Displacement – straight line drawn between
initial position (xi) and final position (xf)
Has both magnitude and direction,
so displacement is a vector quantity.
Don’t use distance and displacement interchangeably.
Your car odometer measures distance.
“As the crow flies” measures displacement.
HW pg 51 #36) A mouse runs straight north 1.414 m,
stops, turn right at 90 and runs another 1.414 m.
Through what distance was it displaced?
xi
1.414 m
1.414 m
v
s = 1.414 2 + 1.414 2 = 2.0m
59. A bumblebee flew 43 m along a twisting path, only
to land on a flower 3.0 m away. If the entire journey
took 10 s, what was its average speed and average velocity?
speed
x
v=
t
velocity
v
v s
v=
t
Ex pg 35) A large clock has a second hand 1.0 m long.
What is its instantaneous velocity at 15 sec past 12:00?
12
9
3
6
Ex pg 35) A large clock has a second hand 1.0 m long.
What is its instantaneous velocity at 15 sec past 12:00?
speed
12
9
3
v=
6
vel
∆x
∆t
Vector Addition
To add vectors, place them head to tail, order doesn’t matter
Vector math means you will either:
1. Plus them, when point same direction
2. Minus them when point exactly opposite directions
3. Pythag them when they are perpendicular
4. Add vector components if they have no special orientation
AND always add them graphically!
1.
+
=
2.
+
=
3.
+
=
4.
+
=
Vector Addition
To add vectors, place them head to tail, order doesn’t matter
Vector math means you will either:
1. Plus them, when point same direction
2. Minus them when point exactly opposite directions
3. Pythag them when they are perpendicular
4. Add vector components if they have no special orientation
AND always add them graphically!
1.
+
=
2.
+
=
3.
+
=
4.
+
=
Vector Addition
To add vectors, place them head to tail, order doesn’t matter
Vector math means you will either:
1. Plus them, when point same direction
2. Minus them when point exactly opposite directions
3. Pythag them when they are perpendicular
4. Add vector components if they have no special orientation
AND always add them graphically!
1.
+
=
2.
+
=
3.
+
=
4.
+
=
Vector Addition
To add vectors, place them head to tail, order doesn’t matter
Vector math means you will either:
1. Plus them, when point same direction
2. Minus them when point exactly opposite directions
3. Pythag them when they are perpendicular
4. Add vector components if they have no special orientation
AND always add them graphically!
1.
+
=
2.
+
=
3.
+
=
4.
+
=
Vector Addition
To add vectors, place them head to tail, order doesn’t matter
Vector math means you will either:
1. Plus them, when point same direction
2. Minus them when point exactly opposite directions
3. Pythag them when they are perpendicular
4. Add vector components if they have no special orientation
AND always add them graphically!
1.
+
=
2.
+
=
3.
+
=
4.
+
=
These vectors need components!
Vector Components
r
A
Ay
A
Ax
Ay
A
Ax
Ay
Formulas:
Ax = A.cosθ
Ay = A.sinθ
Ay
A
Ay
tanθ=
Ay
Ax
Ax
Ex pg 36) A plane flying at 800 km/hr
goes into a 45’ dive. At what speed
does it approach45 the ground?
o
v = 800 km/hr
Ch2 HW#3 p50+
36,38,40,42,53,59,60
Ch2 HW#3 p50+ 36,38,40,42,53,59,60
38. A youngster on a roof of a 19.0 m tall building stands
at the edge and throws a paper plane from a height
of 1.00 m above the roof. It sails around before landing
directly in front of him 15.0 m from the building. What is
the magnitude of the displacement of the plane
from its landing point?
40. After lifting off its launch pad, a rocket is found to be 480 m
directly above an observer who is 360 m due east of the
pad.
What is the displacement of the rocket from the pad at that
moment?
42. A jogger in the city runs 4 blocks north,
2 blocks east, 1 block south,
4 blocks west, 1 block north,
1 block west, and collapses.
Determine the magnitude of
the jogger’s displacement.
53. A cannonball fired from a gun at ground level
located 20 m away from a castle rises high into
the air in a smooth arc and sails down, crashing
into the wall 60 m up from the ground.
Determine the projectile’s displacement.
60. While on a vacation a tourist left the center of town
in a rented car having an odometer reading of 26725.10 km.
He traveled south of west for 6.00 h and then swung
northeast for 14.0 h, ending up 420 km due east of
the center of town. At that point the odometer reading
was 27725.10 km. Compute his average velocity
and average speed.
3 blocks
42. A jogger in the city runs 4 blocks north,
2 blocks east, 1 block south,
4 blocks west, 1 block north,
1 block west, and collapses.
Displacement =
Determine the magnitude of
5 blocks
the jogger’s displacement.
53. A cannonball fired from a gun at ground level
located 20 m away from a castle rises high into
the air in a smooth arc and sails down, crashing
into the wall 60 m up from the ground.
Determine the projectile’s displacement.
60. While on a vacation a tourist left the center of town
in a rented car having an odometer reading of 26725.10 km.
He traveled south of west for 6.00 h and then swung
northeast for 14.0 h, ending up 420 km due east of
the center of town. At that point the odometer reading
was 27725.10 km. Compute his average velocity
and average speed.
4 blocks
Ch 2.4 — Vector Components
Ex1) Vector A is 10 meters long and makes a 30˚ angle with the (+) x-axis.
Vector B is 15 meters long and makes a 45˚ with the (+) x-axis. What is the
resultant vector of these two?
B
A
30o
45o
B
By
45o
A
Ay
Bx
30o
Ax
Ax = A.cosθ
= 10.cos30°
= 8.7 m
Ay = A.sin θ
= 10.sin30°
=5m
Bx = B.cosθ
= 15.cos45°
= 10.6 m
By = B.sin θ
= 15.sin45°
= 10.6 m
By
B
A
Ay
Ax
Bx
Cy
=
B
By
A
+
Ay
Ax
+
Cx = 8.7 + 10.6 = 19.3 m
Cy = 5 + 10.6 = 15.6 m
C=
19.32 + 15.6 2
= 24.9 m
Bx
=
Cx
Ex) A: 10 m 30 degrees to the (-) x-axis
B: 15 m 45 degrees to (+) x-axis
15 m
10 m
30o
45o
Ex) A: 10 m 30 degrees to the (-) x-axis
B: 15 m 45 degrees to (+) x-axis
15 m
45o
10 m
30o
15 m
By
45o
Bx
30
Ay
o
Ax
Ax = A.cosθ
Bx = B.cosθ
= 10.cos30°
= 15.cos45°
= – 8.7 m
= 10.6 m
Ay = A.sin θ
By = B.sin θ
= 10.sin30°
= 15.sin45°
=+5m
= 10.6 m
By
15 m
15 m
By
C
Cy
45o
45o
Bx
30o
Bx
Ay
30
Ax
Ax
Ax = A.cosθ
Bx = B.cosθ
= 10.cos30°
= 15.cos45°
= – 8.7 m
= 10.6 m
Ay = A.sin θ
By = B.sin θ
= 10.sin30°
= 15.sin45°
=+5m
= 10.6 m
Cx
Ax + Bx = Cx
–8.7 + 10.6 = +1.9 m
Ay + By = Cy
5 + 10.6 = +15.6 m
C = 15.7 m
θ = tan −1
Ch2 HW#5 Vectors WS
Ay
o
Cy
Cx
= tan −1
15.6
= 83o
1. 9
Lab 2.1 Velocity
- due Monday
- Ch2 HW#5 is due @ beginning of period
- Ch2 HW#4 Graphing WS due Monday
Ch5 – Practice Graphs
1. Graph of dist v time. Car starts at home,
where is it at: t=2s? ___
t=6s? ___
t=8s? ____
d. Speed at t=1? ___ t=3? ____
t=4.5? ___ t=5.5? ____ t=7? ___
2. Graph of velocity v time.
Velocity at t=1s? ___ t=2.5s? ___
t=3.5? ___ t=4.5s? ____ t=6s? ___
t=7.5s? ___ t=8? ___
What is magnitude of displacement at:
t=2s? ___ t=3s? ___ t=4? ___
t=5s? ___ t=7s? ___ t=8s? ___
dist
(m)
100
90
80
70
60
50
40
30
20
10
1 2 3 4 5 6 7 8 9 10
time (sec)
vel
(m/s)
100
90
80
70
60
50
40
30
20
10
0
-10 1 2 3 4 5 6 7 8 9 10
time (sec)
-20
3. Make 2 graphs: A person walks West (+) at 1 m/s for 2s. She pauses for 1s.
She then continues West at 2 m/s for 2s. She pauses for 1s. She turns
and walks East (–) at 1 m/s for 8s. Graph d vs t , and vel vs t.
6
5
4
3
2
1
d
0
(m) -1
-2
-3
-4
What is her final displacement?
1
2
3
4
5
6
7
8
9
10
11
12 t(sec)-
6
5
4
3
2
1
vel
(m/s)-1
-2
-3
-4
1
2
3
4
5
6
7
8
9
10
11
12 t(sec)-
3. Make 2 graphs: A person walks West (+) at 1 m/s for 2s. She pauses for 1s.
She then continues West at 2 m/s for 2s. She pauses for 1s. She turns
and walks East (–) at 1 m/s for 8s. Graph d vs t , and vel vs t.
6
5
4
3
2
1
d
0
(m) -1
-2
-3
-4
What is her final displacement?
1
2
3
4
5
6
7
8
9
10
11
12 t(sec)-
6
5
4
3
2
1
vel
(m/s)-1
-2
-3
-4
1
2
3
4
5
6
7
8
9
10
11
12 t(sec)-
Ch2 HW#4 Vector WS
1. A catapult fires a boulder with a launch speed of 20m/s at an angle of
60˚. What are the initial components of speed?
v = 20 m/s
30o
2. A skier is moving at 85 km/hr down a ski slope with a 60˚ angle below
the hortiz. At what rate is his altitude decreasing?
1. A catapult fires a boulder with a launch speed of 20m/s at an angle of
60˚. What are the initial components of speed?
v = 20 m/s
30o
vx = v.cosθ
= 20.cos60°
= 10m/s
vy = v.sin θ
= 20.sin60°
= 17m/s
2. A skier is moving at 85 km/hr down a ski slope with a 60˚ angle below
the horizontal. At what rate is his altitude decreasing?
60 o
v = 85km/hr
1. A catapult fires a boulder with a launch speed of 20m/s at an angle of
60˚. What are the initial components of speed?
v = 20 m/s
60o
vx = v.cosθ
= 20.cos60°
= 10m/s
vy = v.sin θ
= 20.sin60°
= 17m/s
2. A skier is moving at 85 km/hr down a ski slope with a 60˚ angle below
the horizontal. At what rate is his altitude decreasing?
60 o
v = 85km/hr
vy = v.sin θ
= 85.sin60°
= 74km/hr
3. A duck is flying northwest (45˚) at 10m/s. What are its north and east
components of velocity?
v = 10 m/s
45o
4. A = 4m at 20˚ to (+x)
B = 3m at 80˚ to (-x)
Find resultant.
4m
20o
3m
80 o
3. A duck is flying northwest (45˚) at 10m/s. What are its north and east
components of velocity?
vx = v.cosθ
= 10.cos45°
= -7m/s (east)
v = 10 m/s
45o
4. A = 4m at 20˚ to (+x)
B = 3m at 80˚ to (-x)
Find resultant.
4m
20o
3m
80 o
vy = v.sin θ
= 10.sin45°
= 7m/s (north)
3. A duck is flying northwest (45˚) at 10m/s. What are its north and east
components of velocity?
vx = v.cosθ
= 10.cos45°
= -7m/s (east)
v = 10 m/s
vy = v.sin θ
= 10.sin45°
= 7m/s (north)
45o
4. A = 4m at 20˚ to (+x)
B = 3m at 80˚ to (-x)
Find resultant.
4m
20o
Ax = 4.cos20°
= +3.76m
Ay = 4.sin20°
= +1.37m
3m
80o
Bx = 3.cos80°
= -0.52m
By = 3.sin80°
= +2.95m
Ax + Bx = Cx
+3.76 + -.52 = +3.24m
Ay + By = Cy
+1.37 + +2.95 = +4.32m
C = 5.4m
θ = tan
−1
Cy
Cx
= tan −1
4.32
= 53o
3.24
Ch2.5 - Relative Motion
Ex.1) A plane flies north at 100m/s in a strong tail wind
of 25m/s north. With what speed does the plane
fly with the respect to the earth?
Ex.2) Plane at 100m/s north, wind at 25m/s south.
What is the speed of plane w.r.t. earth?
Ex.3) Plane at 100m/s north, wind at 25m/s east.
What speed w.r.t. earth?
Ex.4) Plane at 100m/s north.
Wind at 25m/s north-east (45°).
Speed?
93. A hawk 50 m above the ground sees a mouse directly below running due
north at 2.0 m/s. If it reacts immediately, at what angle and speed must
the hawk dive in a straight line, keeping a constant velocity, to intercept its
prey in 5.0 s? Incidentally, the mouse escaped by jumping in a hole.
sy = 50 m
Earth
v=?
v = 2 m/s
Need compatible vectors!
Mouse travels:
sx = vx.t = 2 m/s . 5 sec = 10 m
sy = 50 m
v=?
Hawk travels the hypotenuse.
s = 10 2 + 50 2 = 51m
Earth
v = 2 m/s
Hawk speed and direction:
s 51m
=
10.2 m
s
t
5s
10
θ = tan −1 = 11.3o
50
v=
Ch2 HW#5 p52+ 71,78,81,87,93,95,97
Ch2 HW#5 p52+ 71,78,81,85,87,93,97
71. A toy electric train runs along
a straight length of track. Its
displacement vs time curve is shown.
Is the train’s velocity ever const
and if so, when? What is its inst vel
at t = 2.0 s and at t = 6.5 s?
Did it change direction if so, when?
7
6
5
dist
4
(m)
3
2
1
1
2
3 4 5
time (sec)
1
2
6
7
8
6
5
78. A trolley runs along a straight run
of track the graph is the plot of its
vel vs time. Approx how far did
it travel in the first 3.0 s of its journey?
How far from starting point is it at t = 6.0 s?
speed
4
(m/s)
3
2
1
-1
-2
3
4
5
6 7
time (sec)
81. Each of two runners at either end of a 100m
straight track jogs toward each other at
a constant 5.0 m/s.
How long will it take before they meet?
S = 100m
s1 = 50 m
s2 = 50 m
87. Pennsylvania Avenue and Prince Street intersect Ave
at right angles. If Mary is running at 8.0 m/s NW
toward the intersection along the avenue,
and John is heading NE at 6.0 m/s toward the
John: 6m/s
intersection along the street, what is the speed
of either person wrt each other?
95. At times of flood, the Colorado River reaches
a speed of 48 km/h near the Lava Falls.
Suppose you wanted to cross it there
perpendicularly, in a motorboat capable
of traveling at 48 km/h in still water.
(a) Is it possible? Explain.
(b) If on a quieter day the water flows at 32 km/h,
at what angle would you head the boat to cut
directly across the river?
(c) At what speed with respect to the shore
would you be traveling?
St
Mary: 8m/s
87. Pennsylvania Avenue and Prince Street intersect
at right angles. If Mary is running at 8.0 m/s NW
toward the intersection along the avenue,
and John is heading NE at 6.0 m/s toward the
intersection along the street, what is the speed
of either person wrt each other?
95. At times of flood, the Colorado River reaches
a speed of 48 km/h near the Lava Falls.
Suppose you wanted to cross it there
perpendicularly, in a motorboat capable
of traveling at 48 km/h in still water.
(a) Is it possible? Explain.
(b) If on a quieter day the water flows at 32 km/h,
at what angle would you head the boat to cut
directly across the river?
(c) At what speed with respect to the shore
would you be traveling?
Ave
St
6 m/s
8 m/s
6 m/s
10 m/s
8 m/s
95. At times of flood, the Colorado River reaches
a speed of 48 km/h near the Lava Falls.
Suppose you wanted to cross it there
perpendicularly, in a motorboat capable
of traveling at 48 km/h in still water.
(a) Is it possible? Explain.
(b) If on a quieter day the water flows at 32 km/h,
at what angle would you head the boat to cut
directly across the river?
(c) At what speed with respect to the shore
would you be traveling?
(boat)
48 km/hr
(boat)
48 km/hr
(river)
48 km/hr
(river)
32 km/hr
97. A cruise ship heads 30º W of N at 20 km/h
in a still sea. Someone in a sweatshirt
dashes across the deck traveling
60º E of N at 10 km/h.
What was the jogger’s velocity with respect to the Earth?
ship
20 k/h
30o
60o
runner
10 k/h
Ch 3.1 - Acceleration
7
6
5
vel
4
(m/s) 3
2
1
1
2
3 4 5
time (sec)
6
7
8
Graphing Acceleration
∆y
slope(m) =
∆x
7
6
5
vel
4
(m/s) 3
2
1
1
2
3 4 5
time (sec)
6
7
8
∆v
a=
t
Accl is the slope of the vel-time graph.
Instantaneous Accl - slope of
the line tangent to the curve
at the given time.
7
6
5
vel
4
(m/s) 3
2
1
1
2
3 4 5
time (sec)
6
7
8
aave
∆v v f − vi
=
=
∆t
t
Units:
Since velocity is a vector (speed and direction),
an object can change speed or direction = accl
a ave
∆v v f − vi
=
=
t
∆t
Units :
m
s
2
Since velocity is a vector (speed and direction),
an object accelerates if it changes speed
or
direction
(linear accl)
(circular accl)
Acceleration is a vector
- speed up:
- slow down:
- turn in a circle:
Ex) A robot is traveling at 1.0 m/s along a straight ramp.
If it speeds up to 2.5 m/s in 0.50 sec, what is its accl?
Ch3 HW#1 p79 1,2,3,8,16,18,
19,20,21,27
Ch 3.1 - Average Acceleration
aave
∆v v f − vi
=
=
∆t
t
Since velocity is a vector (speed and direction),
an object can change speed
or
direction
= accl
(linear accl)
(circular accl)
- accl is a vector
Ex) A robot is traveling at 1.0 m/s along a straight ramp.
If it speeds up to 2.5 m/s in 0.50 sec, what is its accl?
vi = 1.0 m/s
vf = 2.5 m/s
t = 0.5 sec
aave
2.5 m − 1.0 m
s
s = 3.0 m
=
s2
0.5 sec
HW #8) Space shuttle goes from a vertical speed of 5.75 m/s at 1.20 sec
to 6.90 m/s at 1.60 sec, while rising 2.30 m.
Determine the average accl.
HW #8) Space shuttle goes from a vertical speed of 5.75 m/s at 1.20 sec
to 6.90 m/s at 1.60 sec, while rising 2.30 m. Determine the average accl.
vi = 5.75 m/s
vf = 6.90 m/s
tf – ti = 1.60 s – 1.20 s = 0.40 s
x = 2.30 m
a=?
(6.90 − 5.75) m
a=
0.40s
s = 2.88 m
HW#16) 2 motorcycle stuntpersons are driving directly at each other
each starting at rest, and each accelerating at an average rate
of 5.5 m/s2. At what speed will they approach each other
2.0 sec into this?
s2
HW#16) 2 motorcycle stuntpersons are driving directly at each other each
starting at rest, and each accelerating at an average rate of 5.5 m/s2.
At what speed will they approach each other 2.0 sec into this?
vi = 0 m/s
a = 5.5 m/s2
t = 2 sec
vf = ?
vi = 0 m/s
a = 5.5 m/s2
t = 2 sec
vf = ?
HW#16) 2 motorcycle stuntpersons are driving directly at each other each
starting at rest, and each accelerating at an average rate of 5.5 m/s2.
At what speed will they approach each other 2.0 sec into this?
vi = 0 m/s
a = 5.5 m/s2
t = 2 sec
vf = ?
5.5 =
vi = 0 m/s
a = 5.5 m/s2
t = 2 sec
vf = ?
vf − 0
2
vf = 11 m/s each, so they approach each other at 22 m/s relative to each other.
Ch3 HW#1 p79+ 1,2,3,8,16,18,19,20,21,27
1. A rocket lifts off its launch pad and travels
straight up attaining a speed of 100 m/s in 10s.
Calculate its average acceleration.
2. A canvasback duck heading south at 50 km/h
at 2:02 am is spotted at 2:06 am still traveling
south but at 40 km/h. Calculate its ave accl
over that interval.
3. An android on guard duty in front of the
Institute of Robotics is heading due south at
1:07 pm at a speed of 10 m/s when it receives
a command to alter course. At 1:09 pm it is
recorded to be moving at 10 m/s due north.
Compute its ave accl over that interval.
HW #8) Space shuttle goes from a vertical speed of 5.75 m/s at 1.20 sec
to 6.90 m/s at 1.60 sec, while rising 2.30 m. Determine the average accl.
vi = 5.75 m/s
vf = 6.90 m/s
tf – ti = 1.60 s – 1.20 s = 0.40 s
x = 2.30 m
a=?
HW#16) 2 motorcycle stuntpersons are driving directly at each other
each starting at rest, and each accelerating at an average rate
of 5.5 m/s2. At what speed will they approach each other
2.0 sec into this?
HW#16) 2 motorcycle stuntpersons are driving directly at each other each
starting at rest, and each accelerating at an average rate of 5.5 m/s2.
At what speed will they approach each other 2.0 sec into this?
vi = 0 m/s
a = 5.5 m/s2
t = 2 sec
vf = ?
5.5 =
vi = 0 m/s
a = 5.5 m/s2
t = 2 sec
vf = ?
vf − 0
2
vf = 11 m/s each, so they approach each other at 22 m/s relative to each other.
18. Fig P18 is a vel-time graph for a test car
on a straight track. The test car initially
40
moved backward in the negative x-direction
30
at 20 m/s. It slowed, came to a stop, and
then moved off in the positive x-direction
20
at t = 2.0 s.
vel 10
What was its ave accl during each of the (m/s) 0
time intervals 0 to 0.5 s,
-10
1.5 to 2.0 s, and 2.0 to 2.5 s?
-20
What was its instantaneous acceleration
at t = 2.25 s?
19. In Fig P18, what was the car’s instantaneous
accl at t = 3.0 s?
Is the instantaneous acceleration + or –
at t = 3.7 s?
How about at t = 1.1 s?
What is the inst accl of the car at t = 0.25 s?
1
2
3
4
t (sec)
7
20. Fig P20 shows the speed-time curves of
three cyclists traveling a straight course.
What are their respective inst speeds at t = 2 s?
vel
Which if any starts out at t = 2 s with
(m/s)
the greatest inst accl?
6
5
4
1
3
2
3
2
1
1
21. Using fig P20, which cyclist has
the greatest inst accl at each of the following
times: t = 2.1 s, 3.3 s, and 6.5 s?
vel
(m/s)
27. Using fig 3.3, graphically determine the car’s
approx accl at 3.8 s into the run.
2
3 4 5
time (sec)
100
90
80
70
60
50
40
30
20
10
5 10 15 20 25 30 35 40
time (sec)
6
7
8
Ch3.2 - Uniformly Accelerated Motion
Ex 1) A red Jag can screech to a stop from 96.54 km/hr in about 3.7 sec.
Compute its accl, assuming it’s constant.
Ex 1) A red Jag can screech to a stop from 96.54 km/hr in about 3.7 sec.
Compute its accl, assuming it’s constant.
vi = 96 km/hr 26.8 m/s
vf = 0 m/s
t = 3.7 sec
a=?
Ex 1) A red Jag can screech to a stop from 96.54 km/hr in about 3.7 sec.
Compute its accl, assuming it’s constant.
vi = 96 km/hr 26.8 m/s
vf = 0 m/s
t = 3.7 sec
f
a=?
a=
v − vi
t
0 − 26.8 m
=
3.7 s
s = −7.2 m
Ex 2) A bicyclist pedaling along a straight road at 8.0 m/s uniformly
accelerated at +3.00 m/s2 for 3.0 sec. Find her final speed.
s
2
Ex 2) A bicyclist pedaling along a straight road at 8.0 m/s uniformly
accelerated at +3.00 m/s2 for 3.0 sec. Find her final speed.
Ex 2) A bicyclist pedaling along a straight road at 8.0 m/s uniformly accelerated at
+3.00 m/s2 for 3.0 sec. Find her final speed.
vi = 8.0 m/s
a = + 3.0 m/s2
t = 3.0 sec
vf = ?
a=
v f − vi
t
vf = vi + at
vf = 8 m/s + (3 m/s2)(3 s)
vf = 17 m/s
vf
v
vi
How do U find the average of 2 numbers?
vf
v
vi
vave = ½(vi + vf)
Mean Speed Theorem:
t
S = ½(vi + vf)t
Ex 3) A bullet is fired from a gun with a muzzle speed of 330 m/s.
If the barrel is 0.152 m long, how long does it take to travel down the barrel?
Ex 3) A bullet is fired from a gun with a muzzle speed of 330 m/s.
If the barrel is 0.152 m long, how long does it take to travel down the barrel?
vi =0 m/s
vf = 330 m/s
s = 0.152 m
t=?
Ex 4) The Jag can go from rest to 13.4 m/s in 3.8 sec, accelerated uniformly
at 3.54 m/s2. How much runway does it use to achieve this?
Ex 4) The Jag can go from rest to 13.4 m/s in 3.8 sec, accelerated uniformly
at 3.54 m/s2. How much runway does it use to achieve this?
vi = 0 m/s
vf = 13.4 m/s
t = 3.8 s
a = 3.54 m/s2
Ex 4) The Jag can go from rest to 13.4 m/s in 3.8 sec, accelerated uniformly
at 3.54 m/s2. How much runway does it use to achieve this?
vi = 0 m/s
vf = 13.4 m/s
t = 3.8 s
a = 3.54 m/s2
s=?
s = ½(vi + vf)t
s = ½(0 + 13.4)3.8
s = 25.5 m
HW# 41) A good male sprinter can run 100 m in 10 sec.
What is his average speed?
He will typically reach a peak speed of 11 m/s at about 5 sec and slow down
toward the finish.
If the accl is fairly constant for the first 5 sec, how fast will he be going 3 sec
into the race?
HW# 41) A good male sprinter can run 100 m in 10 sec. What is his average
speed?
He will typically reach a peak speed of 11 m/s at about 5 sec and slow down
toward the finish.
If the accl is fairly constant for the first 5 sec, how fast will he be going 3 sec
into the race?
12
10
vel
8
(m/s)
6
4
2
1 2 3 4 5 6 7 8 9 10
time (sec)
12
vf = ?
vi = 0 m/s
t = 3 sec
s = _____
a = _____
10
vel
8
(m/s)
6
4
2
1 2 3 4 5 6 7 8 9 10
time (sec)
12
vf = ?
vi = 0 m/s
t = 3 sec
s = _____
a = _____
10
vel
8
(m/s)
6
4
2
1 2 3 4 5 6 7 8 9 10
time (sec)
t = 5 sec
vi = 0 m/s
vf = 11 m/s
11 m
∆v
s = 2.2 m
a=
=
2
s
5 sec
t
vf = vi + at
= 0 + (2.2m/s2)(3sec)
= 6.6 m/s
Ch3 HW# 2 pg 80+ 32,33,41,42,43
Ch3 HW#2 p 80+ 32,33,41,42,43
32. An R75 maintenance robot on a spaceship
is standing in front of the bathroom when it
begins to move down the straight passage
way. It accelerates at a constant 2.0 m/s2.
Find its speed at the end of 5.0 s.
33. With the previous problem in mind,
how far did the robot travel in 5.0 s?
42. If a van moving at 50.0 km/h uniformly
accelerates
up to 70.0 km/h in 20.0 s, how far along
the straight road will it travel in the process?
43. Supposing that the acceleration of a 1997
Corvette is constant (which it really isn’t)
how much road will it travel in going
from 0 to 60 mph ( i.e., 26.8 m/s) in 4.8 s?
Ch 3.5 The Equations for Const Accl
There are 5 motion variables:
vf, vi, a, s, t
There are 5 equations to use them:
1. vf= vi+at
2. vave = ½(vi +vf )
3. s = vave.t = ½(vi +vf)t
4. s = vi.t + ½at2
5. vf2 = vi2 + 2as
Ex1) A lotus can travel a straight 30.0m
from rest in 3.30s. Assume accl is const - find it!
Ex1) A lotus can travel a straight 30m from rest in 3.30s.
Assume accl is const - find it!
s = ~ 30 m
vi = 0 m/s
s = vit + ½at2
t = 3.30 s
s = vit + ½at2
a=?
30 m = ½a(3.3)2
a = 5.5 m/s2
Ex2) The cheetah can achieve speeds of 113 km/hr. They
have been observed on a straight run bounding from a
standing start to achieve 72 km/h in 2.0s.
a) Find the accl
b) What minimum distance to go from rest to 17.9 m/s.
Ex2) The cheetah can achieve speeds of 113 km/hr.
They have been observed on a straight run bounding
from a standing start to achieve 72 km/h in 2.0s.
a) Find the accl
b) What minimum distance to go from rest to 17.9 m/s.
Ex2) The cheetah can achieve speeds of 113 km/hr. They have been
observed on a straight run bounding from a standing start to achieve
72 km/h in 2.0s.
a) Find the accl
b) What minimum distance to go from rest to 17.9 m/s.
v = 30 m/s
vi = 0 m/s
vf = 20 m/s
t = 2 sec
a=?
b) vi = 0 m/s
vf = 17.9 m/s
s=?
a = 10 m/s2
vf = vi + at
a=
v f − vi
t
20 m − 0 m
s
s = 10 m
=
2
s
2 sec
Ex2) The cheetah can achieve speeds of 113 km/hr. They have been
observed on a straight run bounding from a standing start to achieve
72 km/h in 2.0s.
a) Find the accl
b) What minimum distance to go from rest to 17.9 m/s.
v = 30 m/s
vi = 0 m/s
vf = 20 m/s
t = 2 sec
a=?
b) vi = 0 m/s
vf = 17.9 m/s
s=?
a = 10 m/s2
vf = vi + at
a=
v f − vi
t
20 m − 0 m
s
s = 10 m
=
2
s
2 sec
vf2 = vi2 + 2as
17.92 = 02 + 2(10)s
s = 16 m
Ch 3 HW#3 pg 81 46,47,49,53,55,57,64(70xc)
#53) A driver traveling at 60 km/h sees a chicken crossing the road and slams
on the brakes. Accelerating at -7m/s2, the car stops just in time to run down
the chicken 23m ahead. What was the drivers reaction time?
Ch 3 HW#3 pg 81 46,47,49,53,55,57,64(70xc)
#53) A driver traveling at 60 km/h sees a chicken crossing the road and slams
on the brakes. Accelerating at -7m/s2, the car stops just in time to run down
the chicken 23m ahead. What was the drivers reaction time?
trxn = ?
vi = 16.7 m/s
a = – 7 m/s2
vf = 0 m/s
s = 23 m
v =16.7 m/s vi = 16.7 m/s vf = 0 m/s
a = 0 m/s2
a = –7 m/s2
s1
s1 +
s2
s2 = 23 m
#64) A cowboy on a horse rides up to a moving train traveling at 5.0 km/h along
a straight length of track. After running next to the last car for a while, he
charges ahead toward the engine 100m away, and gets there in 1.10 min.
Determine his accl.
Ch3 HW#3 p81 46,47,49,53,55,57,64(70xc)
#64) A cowboy on a horse rides up to a moving train traveling at 5.0 km/h along a straight
length of track. After running next to the last car for a while, he charges ahead toward
the engine 100m away, and gets there in 1.10 min. Determine his accl.
Train:
vcon = 1.4 m/s
Cowboy:
vi = 1.4 m/s
a=?
#64) A cowboy on a horse rides up to a moving train traveling at 5.0 km/h along a straight
length of track. After running next to the last car for a while, he charges ahead toward
the engine 100m away, and gets there in 1.10 min. Determine his accl.
Train:
vcon = 1.4 m/s
t = 66 sec
Cowboy:
vi = 1.4 m/s
a=?
#64) A cowboy on a horse rides up to a moving train traveling at 5.0 km/h along a straight
length of track. After running next to the last car for a while, he charges ahead toward
the engine 100m away, and gets there in 1.10 min. Determine his accl.
Train:
s=v.t
vcon = 1.4 m/s
= 1.4 m/s . 66 s = 92.4 m
t = 66 sec
Cowboy:
vi = 1.4 m/s
a=?
#64) A cowboy on a horse rides up to a moving train traveling at 5.0 km/h along a straight
length of track. After running next to the last car for a while, he charges ahead toward
the engine 100m away, and gets there in 1.10 min. Determine his accl.
Train:
s=v.t
vcon = 1.4 m/s
= 1.4 m/s . 66 s = 92.4 m
t = 66 sec
Cowboy:
s = 192.4 m
vi = 1.4 m/s
a=?
s = vit + ½at2
192.4 = (1.4)(66) + ½(a)(66)2
a = .217 m/s2
Ch3 HW#3 p81+ 46,47,49,53,55,57,64(70xc)
(Day 10)
Lab 2.2 Acceleration
- due Mon
- Ch3 HW#3 due @ beginning of period
- Ch3 HW#4 Eqns due Mon
Ch3 HW#3 p81 46,47,49,53,55,57,64,70
s = 10 m
46. The driver of a car traveling at 10.0 m/s
vi = 10 m/s
vf = ?
along a straight road depresses the accelerator
a = 2.39 m/s2
and uniformly increases her speed at a rate of
2.39 m/s2 straight toward a wall 10 m away.
At what speed will it crash into the wall?
47. The length of a straight tunnel through
s = 25 m
a mountain is 25.0 m. A cyclist heads
directly toward it, accelerating at a constant
vi = 5 m/s
vf = ?
a = 0.20 m/s2
rate of 0.20 m/s2. If at the instant he enters
the tunnel he is traveling at a speed of 5.00 m/s,
how fast will he be moving as he emerges?
Ch3 HW#3 p81 46,47,49,53,55,57,64,70
s = 25 m
47. The length of a straight tunnel through
vi = 5 m/s
vf = ?
a mountain is 25.0 m. A cyclist heads
a = 0.20 m/s2
directly toward it, accelerating at a constant
vf2 = vi2 + 2as
= 52 + 2(0.20)(25)
rate of 0.20 m/s2. If at the instant he enters
= 5.9 m/s
the tunnel he is traveling at a speed of 5.00 m/s,
how fast will he be moving as he emerges?
s = 290 m
49. A Jaguar in an auto accident in England
in 1960 left the longest recorded skid marks vi = ?
vf = 0
a = -3.9 m/s2
on a public road: an incredible 290 m long.
Assuming an average acceleration of –3.9 m/s2
(that is, -0.4 g), calculate the Jag’s speed
when the brakes locked.
Ch3 HW#3 p81 46,47,49,53,55,57,64,70
55. A swimmer stroking along at a fast 2.2 m/s
ceases all body movement and uniformly
coasts to a dead stop in 10 m. Determine
how far she moved during her 3rd second of
drift.
1st find accl:
2nd find s @ t = 2 sec
s = 10 m
vi = 2.2 m/s
vf = 0
3rd second is from
t = 2 sec – t = 3 sec
3rd find s @ t = 3 sec
57. Two trains heading straight for each other on the same track are 250 m
apart when their engineers see each other and hit the brakes. The
Express, heading west at 96 km/h, slows down, decelerating at an average
of 4 m/s2 while the eastbound Flyer, traveling at 110 km/h, slows down,
decelerating at an average of 3 m/s2. Will they collide?
Flyer
vi = 30.6 m/s
a = - 3 m/s2
Express
vf = 0 vf = 0
vi = 26.7 m/s
a = -4 m/s2
(XC) 70. A motorcycle cop, parked at the side of a highway reading a
magazine, is passed by a woman in a red Ferrari 308 GTS doing 90.0 km/h.
After a few attempts to get his cycle started, the officer roars off 2.00 s later.
At what average rate must he accelerate if 110 km/h is his top speed
and he is to catch her just at the state line 2.00 km away?
Ferrari:
Cop:
Ch3 HW#4 - Kinematic Equations WS
1. What is the velocity of a car that starts at rest,
1.
and accelerates at a constant 5 m/s2 for 5 sec?
2. What is the velocity of a car that starts at rest,
and accelerates at a constant 5 m/s2 for 20 m?
3. What is the displacement of an object that starts
with an initial velocity of 10 m/s and undergoes
a constant acceleration of 2 m/s2 for 4 sec?
4. If a bicyclist accelerates at a steady rate to finish
the sprint at 25 m/s and had an average velocity
of 18 m/s over the sprint, what was his initial vel?
5. An object is dropped from the top of a building 35m
tall. If it accelerates at 9.8 m/s2, neglecting air
friction, how long will it take for it to hit the ground?
6. In the last problem, how fast is the object going
right before it hits the ground?
7. If you are curious about the depth of a mine shaft,
old timers will tell you to drop a rock in the shaft
and time until you hear the rock hit the bottom.
If the acceleration of gravity, g, is 9.8 m/s2 and the
rock falls for 3.5 sec, roughly how deep is the hole?
vf= vi+at
Ch3 HW#4 - Kinematic Equations WS
1. vf= vi+at
1. What is the velocity of a car that starts at rest,
and accelerates at a constant 5 m/s2 for 5 sec?
2. vf2 = vi2 + 2as
2. What is the velocity of a car that starts at rest,
and accelerates at a constant 5 m/s2 for 20 m?
3. s = vi.t + ½at2
3. What is the displacement of an object that starts
with an initial velocity of 10 m/s and undergoes
a constant acceleration of 2 m/s2 for 4 sec?
4. vave = ½(vi +vf )
4. If a bicyclist accelerates at a steady rate to finish
the sprint at 25 m/s and had an average velocity
of 18 m/s over the sprint, what was his initial vel?
5. s = vi.t + ½at2
5. An object is dropped from the top of a building 35m
tall. If it accelerates at 9.8 m/s2, neglecting air
friction, how long will it take for it to hit the ground?
6. vf= vi+at
6. In the last problem, how fast is the object going
right before it hits the ground?
7. If you are curious about the depth of a mine shaft,
old timers will tell you to drop a rock in the shaft
7. s = vi.t + ½at2
and time until you hear the rock hit the bottom.
If the acceleration of gravity, g, is 9.8 m/s2 and the
rock falls for 3.5 sec, roughly how deep is the hole?
Ch 3.5 - Acceleration Due To Gravity
Strato (300 B.C.) — objects in free fall are accelerating.
He recognized this with rain drops.
Galileo (1500’s) — With no precise measuring devices,
“diluted” gravity using inclined planes.
No way to measure final velocity so derived:
s =vit + ½at2
(s is proportional to t2)
double time quadruple displacement
g = 9.8m/s2 (average at earth’s surface)
Up and Down
Doesn’t matter if object goes up, down, or sideways,
gravity pulls down on it at 9.81m/s2.
Ex1) A salmon is dropped by a hovering eagle.
How far will it fall in 2.5s? (No air drag)
Ex p66) A salmon is dropped by a hovering eagle.
How far will it fall in 2.5s? (No air drag)
t = 2.5 sec
s=?
Ex p66) A salmon is dropped by a hovering eagle.
How far will it fall in 2.5s? (No air drag)
t = 2.5 sec
s=?
a = 9.8 m/s2
vi = 0
Ex p66) A salmon is dropped by a hovering eagle.
How far will it fall in 2.5s? (No air drag)
t = 2.5 sec
s=?
a = 9.8 m/s2
vi = 0
s = vit + ½at2
= 0 + ½(9.8)(2.5)2
= 31 m
Ex2) A ball is thrown straight down
from a roof at 10.0m/s.
If the building is 100m tall, what speed hit ground?
How long did the trip take?
vi = 10 m/s
s = 100 m
a = 9.8 m/s2
vf = ?
t=?
Ex p66) A ball is thrown straight down
from a roof at 10.0m/s.
If the building is 100m tall, what speed hit ground?
How long did the trip take?
vi = 10 m/s
s = 100 m
a = 9.8 m/s2
vf = ?
t=?
vf2 = vi2 + 2as
vf = vi + at
vf = 45 m/s
t = 3.5 sec
Free Fall
At Earth’s surface, all objects falling thru a vacuum
accelerate at the same constant rate,
regardless of weight, shape, etc.
Most people believe falling bodies descends at a rate
proportional to their weight.
Galileo started our correct thinking
Air Drag (contribute to our incorrect thinking)
As an object falls faster, breaks thru more air,
creating a larger resistance force.
Eventually reach the point where
force resisting = force pulling
Fair = Fg
If the net force = 0 the object will continue with
a constant velocity
Terminal speed – depends on
shape, surface area, and weight of object.
a = 9.8 m/s2
vi = 0
Fg
Free Fall
At Earth’s surface, all objects falling thru a vacuum
accelerate at the same constant rate,
regardless of weight, shape, etc.
Most people believe falling bodies descends at a rate
proportional to their weight.
Galileo started our correct thinking
Air Drag (contribute to our incorrect thinking)
As an object falls faster, breaks thru more air,
creating a larger resistance force.
Eventually reach the point where
force resisting = force pulling
Fair = Fg
If the net force = 0 the object will continue with
a constant velocity
Terminal speed – depends on
shape, surface area, and weight of object.
Fr
v
a = less
Fg
Free Fall
At Earth’s surface, all objects falling thru a vacuum
accelerate at the same constant rate,
regardless of weight, shape, etc.
Most people believe falling bodies descends at a rate
proportional to their weight.
Galileo started our correct thinking
Air Drag (contribute to our incorrect thinking)
As an object falls faster, breaks thru more air,
creating a larger resistance force.
Eventually reach the point where
force resisting = force pulling
Fair = Fg
If the net force = 0 the object will continue with
a constant velocity
Terminal speed – depends on
shape, surface area, and weight of object.
Fr
vmax
a=0
Fg
Ch3 HW#4 p82+ 71,72,73,74,75
HW #71. EXPLORING PHYSICS ON YOUR OWN: ask a friend to hold
his or her thumb and forefinger parallel to each other in a horizontal
plane. The fingers should be about an inch apart. Now you hold a 1ft ruler vertically in the gap just above and between them. Have your
friend look at the ruler and catch it when you, without warning, let it
fall. Calculate the corresponding response time. Now position a
dollar bill vertically so that Washington’s face is between your
friend’s fingers
- is it likely to be caught when dropped?
72. A kangaroo can jump straight up about 2.5 m.
What is it’s takeoff speed?
s = 2.5 m
vf = 0
a = - 9.8 m/s2
vi = ?
72. A kangaroo can jump straight up about 2.5 m.
What is it’s takeoff speed?
s = 2.5 m
vf = 0
a = - 9.8 m/s2
vi = ?
vf2 = vi2 + 2as
0 = vi2 + 2(-9.8)(2.5)
vi = 7 m/s
(Day 12)
Lab 3.1 Free Fall
- due 2 days
- Ch3 HW#5 due @ beginning of period
- Ch3 HW#6 Gravity WS due tomorrow
Ch3 HW#5 p82+ 71 – 75
71. EXPLORING PHYSICS ON YOUR OWN: ask a friend to
hold his or her thumb and forefinger parallel to each other in
a horizontal plane. The fingers should be about an inch
apart. Now you hold a 1-ft ruler vertically in the gap just
above and between them. Have your friend look at the ruler
and catch it when you, without warning, let it fall. Calculate
the corresponding response time. Now position a dollar bill
vertically so that Washington’s face is between your friend’s
fingers
- is it likely to be caught when dropped?
72. A kangaroo can jump straight up about 2.5 m.
What is it’s takeoff speed?
s = 2.5 m
vf = 0
a = - 9.8 m/s2
vi = ?
72. A kangaroo can jump straight up about 2.5 m.
What is it’s takeoff speed?
s = 2.5 m
vf = 0
a = - 9.8 m/s2
vi = ?
vf2 = vi2 + 2as
0 = vi2 + 2(-9.8)(2.5)
vi = 7 m/s
Ch3 HW#5 p82+ 71 – 75
73. At what speed would you hit the floor
if you stepped off a chair 0.50 m high? Ignore friction.
Ch3 HW#5 p82+ 71 – 75
74. If a stoned dropped (not thrown) from a bridge takes
3.7 s to hit the water, how high is the rock-dropper?
Ignore friction.
s = vit + ½at2
Ch3 HW#5 p82+ 71 – 75
75. Ignoring air fiction, how fast will an object be moving
and how far will it have fallen after dropping from rest
for 1.0 s
vf
s
2.0 s
5.0 s
vf = vi + at
s = vit + ½at2
10 s?
Ch3 HW#6 Free Fall Worksheet
1. An object is dropped from a height of 50 m.
If air resistance is negligible, how long will it
take to hit the ground?
2. An object is dropped from a height of 100 m.
If air resistance is negligible, how long will it
take to hit the ground?
3. How do the times compare when dropped
from 50 m and 100 m?
4. An object is dropped from an unknown height
and falls for 1.3 sec. If air resistance is
negligible, how tall is the height of the drop?
5. An object is dropped from an unknown height
and falls for 2.6 sec. If air resistance is
negligible, how tall is the height of the drop?
6. How do the heights differ when the time
doubles?
Ch 3.6 Up and Down
If a ball is thrown upward, its speed diminishes by 9.81 m/s every second.
Momentarily stops at the top - peak altitude highest vertical distance.
The fall from the top is the same as if it were dropped from that height.
Ex1) A bullet has a muzzle speed of 200 m/s.
If it is shot straight up, what is peak altitude?
- how fast does it come back to the shooter?
- how long does the whole trip take?
s=?
vi = 200 m/s vf = ?
t=?
Ch 3.6 Up and Down
If a ball is thrown upward, its speed diminishes by 9.81 m/s every second. Momentarily
stops at the top - peak altitude highest vertical distance.
The fall from the top is the same as if it were dropped from that height.
Ex p67) A bullet has a muzzle speed of 200 m/s.
If it is shot straight up, what is peak altitude?
- how fast does it come back to the shooter?
- how long does the whole trip take?
a = - 9.8 m/s2
vi = 200 m/s
vf = ?
t=?
Ch 3.6 Up and Down
If a ball is thrown upward, its speed diminishes by 9.81 m/s every second. Momentarily
stops at the top - peak altitude highest vertical distance.
The fall from the top is the same as if it were dropped from that height.
Ex p67) A bullet has a muzzle speed of 200 m/s.
If it is shot straight up, what is peak altitude?
- how fast does it come back to the shooter?
- how long does the whole trip take?
Split the problem on half, solve the upward path.
vf = 0
ttop = ?
a = - 9.8 m/s2
vi = 200 m/s
vf = ?
t=?
Ch 3.6 Up and Down
If a ball is thrown upward, its speed diminishes by 9.81 m/s every second. Momentarily
stops at the top - peak altitude highest vertical distance.
The fall from the top is the same as if it were dropped from that height.
Ex p67) A bullet has a muzzle speed of 200 m/s.
If it is shot straight up, what is peak altitude?
- how fast does it come back to the shooter?
- how long does the whole trip take?
Split the problem on half, solve the upward path.
vf = 0
s = ? ttop = ?
vfbottom = 200 m/s
vf = vi + attop
a = - 9.8 m/s2
0 = 200 + (-9.8)t
vi = 200 m/s
vf = ?
t=?
ttop = 20 sec
ttotal = 40 sec
vf2 = vi2 + 2as
0 = 2002 + 2(-9.8)s
s = 2000 m
Ex2) A ball is hurled straight up at a speed of 15.0 m/s leaving the hand of the
thrower 2.00 m above the ground. Compute the times and the balls speed
when it passes an observer sitting in a window in line with the throw 10.0 m
above the point of release.
s = 10 m
vi = 15 m/s
s=2m
Ex pg 68) A ball is hurled straight up at a speed of 15.0 m/s leaving the hand of
the thrower 2.00 m above the ground. Compute the times and the balls
speed when it passes an observer sitting in a window in line with the throw
10.0 m above the point of release.
vf = 0 t = ?
a = - 9.8 m/s2
s = 10 m
vi = 15 m/s
s=2m
Ex pg 68) A ball is hurled straight up at a speed of 15.0 m/s leaving the hand of
the thrower 2.00 m above the ground. Compute the times and the balls
speed when it passes an observer sitting in a window in line with the throw
10.0 m above the point of release.
vf = 0 t = ?
vup = ? tup = ?
vdown =? tdown = ?
a = - 9.8 m/s2
s = 10 m
vi = 15 m/s
s=2m
Ex pg 68) A ball is hurled straight up at a speed of 15.0 m/s leaving the hand of
the thrower 2.00 m above the ground. Compute the times and the balls
speed when it passes an observer sitting in a window in line with the throw
10.0 m above the point of release.
vf = 0 t = ?
vup = ? tup = ?
vdown =? tdown = ?
vup:
vf2 = vi2 + 2as
a = - 9.8 m/s2
s = 10 m
= 152 + 2(-9.8)(10)
vi = 15 m/s
= 5.4 m/s
s=2m
vdown = 5.4 m/s too.
Ex pg 68) A ball is hurled straight up at a speed of 15.0 m/s leaving the hand of
the thrower 2.00 m above the ground. Compute the times and the balls
speed when it passes an observer sitting in a window in line with the throw
10.0 m above the point of release.
vf = 0 t = ?
vup = ? tup = ?
vdown =? tdown = ?
vup:
vf2 = vi2 + 2as
a = - 9.8 m/s2
s = 10 m
= 152 + 2(-9.8)(10)
vi = 15 m/s
= 5.4 m/s
s=2m
vdown = 5.4 m/s too.
tup:
To find tdown need ttop:
vup = vi + atup
vf = vi + attop
ttotal = 3.06 sec
5.4 = 15 + (-9.8)t
0 = 15 + (-9.8)ttop
tdown = 3.06 – 0.98 = 2.08 sec
tup = 0.98 sec
ttop = 1.53
Ch3 HW#7 pg 82+ 76,77,81,82,88,89
#82) An arrow is launched vertically upward from a crossbow at 98.1 m/s.
a) What is the inst speed at the end of 10 sec?
b) What is its ave speed to that moment?
c) How high has it risen?
d) What is the inst accl 4.10 sec into flight?
t = 10 sec
vf = ? s = ?
a = - 9.8 m/s2
vf = 98.1 m/s
Ch3 HW#7 pg 82+ 76,77,81,82,88,89
#82) An arrow is launched vertically upward from a crossbow at 98.1 m/s.
a) What is the inst speed at the end of 10 sec?
b) What is its ave speed to that moment?
c) How high has it risen?
d) What is the inst accl 4.10 sec into flight?
a) vf = vi + at
= 98.1 + (-9.8)(10)
t = 10 sec
vf = ? s = ?
= 0 m/s
(at the top of its path)
a = - 9.8
m/s2
vf = 98.1 m/s
Ch3 HW#7 pg 82+ 76,77,81,82,88,89
#82) An arrow is launched vertically upward from a crossbow at 98.1 m/s.
a) What is the inst speed at the end of 10 sec?
b) What is its ave speed to that moment?
c) How high has it risen?
d) What is the inst accl 4.10 sec into flight?
a) vf = vi + at
t = 10 sec
vf = ? s = ?
= 98.1 + (-9.8)(10)
= ½(98.1 + 0)
= 0 m/s
= 49 m/s
(at the top of its path)
a = - 9.8
m/s2
vf = 98.1 m/s
b) vave = ½(vi + vf)
Ch3 HW#7 pg 82+ 76,77,81,82,88,89
88) Imagine someone dropped a firecracker off a roof and heard it explode
10 sec later. If the speed of sound is 330 m/s, calculate how far down
it was when it exploded.
vi = 0 m/s
a = 9.8 m/s2
tdown = ?
v = 330 m/s
tup = ?
s=?
Ch3 HW#7 pg 82+ 76,77,81,82,88,89
88) Imagine someone dropped a firecracker off a roof and heard it explode 10 sec later.
If the speed of sound is 330 m/s, calculate how far down it was when it exploded.
vi = 0 m/s
a = 9.8 m/s2
tdown = ?
tup + tdown = 10 sec
v = 330 m/s
tup = ?
s=?
sup = sdown
Ch3 HW#7 pg 82+ 76,77,81,82,88,89
88) Imagine someone dropped a firecracker off a roof and heard it explode 10 sec later.
If the speed of sound is 330 m/s, calculate how far down it was when it exploded.
vi = 0 m/s
tup + tdown = 10 sec
sup = sdown
sdown = vit + ½atdown2 sup = v.tup
sdown = 4.9tdown2
a = 9.8
m/s2
tdown = ?
sup = 330tup
v = 330 m/s
4.9tdown2 = 330tup
tup = ?
4.9tdown2 = 330(10 – tdown)
tdown = 8.8 sec
s=?
(Day 14)
Ch3 HW#7 pg 82+ 76,77,81,82,88,89
76. A cannonball is fired straight up at a rather modest speed of 9.8m/s. Compute its
maximum altitude and the time it takes to reach that height.
77. Calculate the speed at which hailstone, falling from 0.9144x104 m out of a
cumulonimbus cloud, would strike the ground, presuming air friction is negligible (which it
certainly is not.) Give your answer in m/s.
81. A young kid with a huge baseball cap is playing catch with himself by throwing
a ball straight up. How fast does he throw if the ball comes back to his hands
a second later? At low speeds air friction is negligible.
ttop = 0.5 sec
vf = 0 m/s
a = - 9.8 m/s2
vi = ?
ttotal = 1 sec
89. A bag of sand dropped by a would-be assassin from the roof of a building just
misses Tough Tony, a gangster 2 m tall.
The missile traverses the height of Tough Tony in 0.20 s, landing with a thud at his
feet. How high was the building? Ignore friction.
vi = 0 m/s
a = 9.8 m/s2
s=?
s=2m
t = 0.20 sec
89. A bag of sand dropped by a would-be assassin from the roof of a building just
misses Tough Tony, a gangster 2 m tall.
The missile traverses the height of Tough Tony in 0.20 s, landing with a thud at his
feet. How high was the building? Ignore friction.
vi = 0 m/s
a = 9.8 m/s2
s=?
Use Tony’s givens to find the speed near his head:
s = vit + ½at2
s=2m
t = 0.20 sec
2 = vi(.2) + ½(9.8)(.2)2
vi = 9.02 m/s
89. A bag of sand dropped by a would-be assassin from the roof of a building just
misses Tough Tony, a gangster 2 m tall.
The missile traverses the height of Tough Tony in 0.20 s, landing with a thud at his
feet. How high was the building? Ignore friction.
vi = 0 m/s
a = 9.8 m/s2
s=?
Now find the final speed:
vi = 9.02 m/s
s=2m
t = 0.20 sec
vf = ?
vf = vi + at
vf = 9.02 + 9.8(.2)
vf = 10.98 m/s
89. A bag of sand dropped by a would-be assassin from the roof of a building just
misses Tough Tony, a gangster 2 m tall.
The missile traverses the height of Tough Tony in 0.20 s, landing with a thud at his
feet. How high was the building? Ignore friction.
vi = 0 m/s
Now the height of the building is easy:
a = 9.8 m/s2
s=?
vf = 10.98 m/s
Ch 3.7 – Projectile Motion
vi
An object shot at an angle
has two motions that are
independent of each other.
Ch 3.7 – Projectile Motion
t=1s
t=2s
t=3s
t=4s
An object shot at an angle
has two motions that are
independent of each other.
Vertical motion controlled by gravity
vi
t=1s
t=2s
t=3s
t=4s
v1 = vi
v2 = vi
v3 = vi
v4 = vi
An object shot at an angle
has two motions that are
independent of each other.
Vertical motion controlled by gravity
Horizontal motion controlled
by the initial velocity
given to the projectile.
vi
t=1s
t=2s
t=3s
t=4s
t=1s
t=2s
t=3s
t=4s
v1 = vi
v2 = vi
v3 = vi
v4 = vi
An object shot at an angle
has two motions that are
independent of each other.
vertical motion controlled by gravity
Horizontal motion controlled
by the initial velocity
given to the projectile.
vi
t=1s
t=2s
t=3s
t=1s
t=2s
t=3s
t=4s
v1 = vi
v2 = vi
v3 = vi
v4 = vi
An object shot at an angle
has two motions that are
independent of each other.
Vertical motion controlled by gravity
Horizontal motion controlled
by the initial velocity
given to the projectile.
t=4s
Only thing that links the 2 motions is time
vi
t=1s
t=2s
t=3s
t=4s
t=1s
t=2s
t=3s
t=4s
v1 = vi
v2 = vi
v3 = vi
v4 = vi
An object shot at an angle
has two motions that are
independent of each other.
Vertical motion controlled by gravity
Horizontal motion controlled
by the initial velocity
given to the projectile.
Ex pg70) a youngster hurls a ball horizontally
at a speed of 10 m/s from a bridge 50 m
above a river.
a) How long will it take to hit the water?
b) What is velocity just before it hits the water?
c) How far from the bridge will it hit?
vi = 10 m/s
sy = 50 m
A) The ball will hit the ground
at the same time, whether
It is thrown horizontally
or dropped. So solve for
the dropped ball!
vi = 10 m/s
sy = 50 m
A) The ball will hit the ground
at the same time, whether
It is thrown horizontally
or dropped. So solve for
the dropped ball!
vi = 10 m/s
sy = 50 m
sy = vit + ½at2
50 = 0 + ½(9.8)t2
t = 3.2 sec
vi = 10 m/s
sy = 50 m
sy = vit + ½at2
50 = 0 + ½(9.8)t2
t = 3.2 sec
A) The ball will hit the ground
at the same time, whether
It is thrown horizontally
or dropped. So solve for
the dropped ball!
C) If there were no gravity,
the ball would travel
the same distance away
from the cliff, as the
actual projectile.
vi = 10 m/s
A) The ball will hit the ground
at the same time, whether
It is thrown horizontally
or dropped. So solve for
the dropped ball!
C) If there were no gravity,
the ball would travel
the same distance away
from the cliff, as the
actual projectile.
sx = vit + ½at2
= vcont + 0
= 10 m/s(3.2 s)
= 32 m
sy = 50 m
sy = vit + ½at2
50 = 0 + ½(9.8)t2
t = 3.2 sec
vi = 10 m/s
A) The ball will hit the ground
at the same time, whether
It is thrown horizontally
or dropped. So solve for
the dropped ball!
C) If there were no gravity,
the ball would travel
the same distance away
from the cliff, as the
actual projectile.
sx = vit + ½at2
= vcont + 0
= 10 m/s(3.2 s)
= 32 m
sy = 50 m
B) Find both components
of final velocity, and pythag!
sy = vit + ½at2
50 = 0 + ½(9.8)t2
t = 3.2 sec
A) The ball will hit the ground
at the same time, whether
It is thrown horizontally
or dropped. So solve for
the dropped ball!
C) If there were no gravity,
the ball would travel
the same distance away
from the cliff, as the
actual projectile.
vi = 10 m/s
sx = vit + ½at2
= vcont + 0
= 10 m/s(3.2 s)
= 32 m
sy = 50 m
B) Find both components
of final velocity, and pythag!
vfx
sy = vit + ½at2
50 = 0 + ½(9.8)t2
t = 3.2 sec
vfy
vf
vfx = 10 m/s
vfy = viy + at
= 0 + (9.8)(3.2)
= 31 m/s
v f = 10 2 + 312 = 32.6m / s
Ch3 HW#8 p82+ 90,92,93,99
92. Suppose you point a rifle horizontally directly at the center of the paper
target 100 m away from you. If the muzzle speed of the bullet is 100 m/s,
where will it strike the target?
vi = 100 m/s
sy = ?
sx = ?
Ch3 HW#8 p82+ 90,92,93,99
92. Suppose you point a rifle horizontally directly at the center of the paper target
100 m away from you. If the muzzle speed of the bullet is 100 m/s, where will it
strike the target?
vi = 100 m/s
sy = ?
sx = ?
sx = vixt + ½at2
100 = 100t + 0
t = 1 sec
Ch3 HW#8 p82+ 90,92,93,99
90. (Modified) A shoe is flung horizontally at 6.0 m/s, and hits the ground 2.0 sec
later. What vertical height was it thrown from? What horizontal distance will it
travel?
vi = 6 m/s
sy = ?
t = 2sec
sx = ?
Ch3 HW#8 p82+ 90,92,93,99
90. (Modified) A shoe is flung horizontally at 6.0 m/s, and hits the ground 1.0 sec
later. What vertical height was it thrown from? What horizontal distance will it
travel?
sy = viyt + ½at2
vi = 6 m/s
sx = vixt
sy = ?
t = 1sec
sx = ?
93. A raw egg is thrown horizontally straight out of the open window
of a fraternity house. If its initial speed is 20 m/s and it hits ground
2.0 s later, at what height was it launched?
vix = 20 m/s
sy = ?
t = 2 sec
93. A raw egg is thrown horizontally straight out of the open window of a fraternity house.
If its initial speed is 20 m/s and it hits ground 2.0 s later, at what height was it launched?
vix = 20 m/s
sy = viyt + ½at2
sy = ?
t = 2 sec
99. Two diving platforms 10-m high terminate just at the edge of each end of a
swimming pool 30-m long. How fast must two clowns run straight off their respective
boards if they are to collide at the surface of the water midpool?
vix = ?
vix = ?
sy = 10 m
sy = 10 m
s = 30 m
99. Two diving platforms 10-m high terminate just at the edge of each end of a
swimming pool 30-m long. How fast must two clowns run straight off their respective
boards if they are to collide at the surface of the water midpool?
vix = ?
vix = ?
sy = 10 m
sy = 10 m
sy = viyt + ½at2
10 = 0 + ½(9.8)t2
s = 15 m
99. Two diving platforms 10-m high terminate just at the edge of each end of a
swimming pool 30-m long. How fast must two clowns run straight off their respective
boards if they are to collide at the surface of the water midpool?
vix = ?
vix = ?
sy = 10 m
sy = 10 m
sy = viyt + ½at2
10 = 0 +
sx = vx.t
½(9.8)t2
s = 15 m
Ch3.7 – Projectiles at an Angle
Ex1) My tennis ball cannon can launch tennis balls at 50 m/s.
o
If I angle it at 40 to the horizontal, what distance can I get?
vi = 50 m/s
40o
sx = ?
Ch3.7 – Projectiles at an Angle
Ex) My tennis ball cannon can launch tennis balls at 50 m/s.
o
If I angle it at 40 to the horizontal, what distance can I get?
vi = 50 m/s
40o
viy
sx = ?
40o
vix
vix = vi ⋅ cos θ
viy = vi ⋅ sin θ
vix = 50 m ⋅ cos 40o
s
vix = 38 m
s
viy = 50 m ⋅ sin 40o
s
viy = 32 m
s
Ch3.7 – Projectiles at an Angle
Ex) My tennis ball cannon can launch tennis balls at 50 m/s.
o
If I angle it at 40 to the horizontal, what distance can I get?
vfy = 0 m/s
ttop = ?
vi = 50 m/s
40
a = -9.8m/s
o
sx = ?
viy = 32 m/s
40o
vix
sx = vx.ttotal
vix = vi ⋅ cos θ
viy = vi ⋅ sin θ
vf = vi + attop
vix = 50 m ⋅ cos 40o
s
vix = 38 m
s
viy = 50 m ⋅ sin 40o
s
viy = 32 m
s
0 = 32 + (-9.8)ttop
= (38 m/s)(6.6 sec)
ttop = 3.3 sec
= 254 m
ttotal = 6.6 sec
Ex2) A football is kicked by a punter with an initial speed of 30 m/s
at 45°, how far will it travel?
Ex2) A football is kicked by a punter with an initial speed of 30 m/s at 45°,
How far will it travel?
vix = 21.2 m/s
viy = 21.2 m/s
ttop = 2.16 s
ttotal = 4.3 s
sx = 91m
Ch3 HW#9 1 – 3
Lab 3.2 – Projectile Motion
- due @ end of period
- Ch3 HW#9 Go over @ beginning of period
- Ch3 HW#10 due tomorrow
Ch3 HW#9 1 – 3
1) A cannonball is fired at 100 m/s at 30°. If air resistance is negligible,
how much time elapses until it hits the ground
at the same height it was launched?
vi = 100 m/s
30 o
Ch3 HW#9 1 – 3
1) A cannonball is fired at 100 m/s at 30°. If air resistance is negligible,
how much time elapses until it hits the ground
at the same height it was launched?
2) How far does it travel?
v = 0 m/s
fy
ttop = ?
a = -9.8m/s
viy = 50 m/s
30o
vix
Ch3 HW#9 1 – 3
1) A cannonball is fired at 100 m/s at 30°. If air resistance is negligible,
how much time elapses until it hits the ground
at the same height it was launched?
2) How far does it travel?
v = 0 m/s
s = vx.t
= (87m/s)(10s)
= 870m
vix = 87 m
fy
ttop = ?
a = -9.8m/s
viy = 50 m/s
s
viy = 50 m
s
30o
vix
3) A football is kicked by a punter with an initial speed of 25 m/s at 40°,
How far will it travel?
vfy = ___m/s
vi = 25 m/s
ttop = ?
40 o
sx = ?
a = ___m/s
viy = ___m/s
40 o
vix
3) A football is kicked by a punter with an initial speed of 25 m/s at 40°,
How far will it travel?
vfy = ___m/s
vi = 25 m/s
ttop = ?
40 o
vix = vi ⋅ cos θ
sx = ?
vix = 25 m ⋅ cos 40 o
s
vix = 19 m
s
viy = vi ⋅ sin θ
viy = 25 m ⋅ sin 40o
s
viy = 16 m
s
a = ___m/s
viy = ___m/s
vf = vi + a.ttop
40 o
vix
3) A football is kicked by a punter with an initial speed of 25 m/s at 40°,
How far will it travel?
vfy = 0 m/s
vi = 25 m/s
ttop = ?
40 o
sx = ?
a = -9.8m/s
viy = 16 m/s
vix = 19 m
s
viy = 16 m
s
40 o
vix
3) A football is kicked by a punter with an initial speed of 25 m/s at 40°,
How far will it travel?
vfy = 0 m/s
vi = 25 m/s
ttop = ?
40 o
sx = ?
a = -9.8m/s
viy = 16 m/s
vix = 19 m
s
viy = 16 m
s
40 o
vix
Ch11 HW#10 4 – 7
4. Cannonball fired at 50 m/s straight up, how much time and how far?
vfy = 0 m/s
vi = 50m/s
θ
t = ? sx = ?
ttop = ?
a = -9.8m/s
viy = ___m/s
viy = vi ⋅ sin θ vix = vi ⋅ cos θ
vf = vi + a.ttop
0 = viy + (-9.8)ttop
vix
sx = vx.ttotal
4. Cannonball fired at 50 m/s straight up, how much time and how far?
5. Cannonball fired at 50 m/s at 30°, how much time and how far?
vfy = 0 m/s
vi = 50m/s
θ
t = ? sx = ?
ttop = ?
a = -9.8m/s
viy = vi ⋅ sin θ
vix = vi ⋅ cos θ
viy 4 = 50 m
vix = 0
s
viy = ___m/s
vix
5. Cannonball fired at 50 m/s at 30°, how much time and how far?
6. Cannonball fired at 50 m/s at 60°, how much time and how far?
vfy = 0 m/s
vi = 50m/s
θ
ttop = ?
t = ? sx = ?
a = -9.8m/s
viy = vi ⋅ sin θ
vix = vi ⋅ cos θ
viy 5 = 25 m
vix = 43 m
s
s
viy = ___m/s
vix
6. Cannonball fired at 50 m/s at 60°, how much time and how far?
7. Cannonball fired at 50 m/s at 45°, how much time and how far?
vfy = 0 m/s
vi = 50m/s
θ
ttop = ?
t = ? sx = ?
a = -9.8m/s
viy = ___m/s
viy = vi ⋅ sin θ
vix = vi ⋅ cos θ
viy 6 = 43 m
vix = 25 m
s
s
vix
7. Cannonball fired at 50 m/s at 45°, how much time and how far?
vfy = 0 m/s
vi = 50m/s
θ
ttop = ?
t = ? sx = ?
a = -9.8m/s
viy = ___m/s
viy = vi ⋅ sin θ
vix = vi ⋅ cos θ
viy 7 = 35 m
vix = 35 m
s
s
vix
4. Cannonball fired at 50 m/s straight up, how much time and how far?
5. Cannonball fired at 50 m/s at 30°, how much time and how far?
6. Cannonball fired at 50 m/s at 60°, how much time and how far?
7. Cannonball fired at 50 m/s at 45°, how much time and how far?
vfy = 0 m/s
vi = 50m/s
θ
t = ? sx = ?
ttop = ?
a = -9.8m/s
viy = ___m/s
vix
Ch3.8 More Projectiles
Ex1) A ball is thrown at 15 m/s at an angle of 20°below the horizontal,
from a cliff that is 100m tall. Where does the ball hit, and what is its final speed?
Ex1) A ball is thrown at 15 m/s at an angle of 20°below the horizontal,
from a cliff that is 100m tall. Where does the ball hit, and what is its final speed?
vi = 15 m/s
o
vix = 15 m ⋅ cos 20 o viy = 15 m ⋅ sin 20
s
s
vix = 14.1 m
sx = ?
s
viy = 5.1 m
s
Ex1) A ball is thrown at 15 m/s at an angle of 20°below the horizontal,
from a cliff that is 100m tall. Where does the ball hit, and what is its final speed?
vi = 15 m/s
vix = 15 m ⋅ cos 20 o
s
vix = 14.1 m
s
viy = 15 m ⋅ sin 20 o
s
viy = 5.1 m
s
sy = viyt + ½at2
100 = 5.1.t + ½(9.8)t2
t = 4.03s
sx = ?
Ex1) A ball is thrown at 15 m/s at an angle of 20°below the horizontal,
from a cliff that is 100m tall. Where does the ball hit, and what is its final speed?
vi = 15 m/s
vix = 15 m ⋅ cos 20 o
s
vix = 14.1 m
s
sy = viyt + ½at2
100 = 5.1.t + ½(9.8)t2
t = 4.03s
sx = ?
viy = 15 m ⋅ sin 20o
s
viy = 5.1 m
s
sx = vx.ttotal
= (14.1m/s)(4.03s)
= 56.8 m
vfy = viy + a.t
= 5.1+ (9.8)(4.03)
vfy = 44.6 m/s
v f = 44.6 2 + 14.12 = 46.8m / s
Ex2) A golf ball is hit at 50m/s at angle of 45˚ above the horizontal.
Neglecting air resistance,
a. How fast is it going as it impacts the ground?
b. Where is its speed a maximum?
c. Where is its speed a minimum?
Justify your answers.
Ex3) A golf ball is hit at 50m/s at angle of 45˚ above the horizontal.
The ball hits a tree, a horizontal distance of 122.5m away from the tee box.
How high up the tree did it hit? How fast was is moving when it struck the tree?
(Use a = 10m/s2)
(It’s a tall redwood!)
sy = ?
vfinal = ?
vi = 50m/s 45°
sx = 200m
Ex3) A golf ball is hit at 50m/s at angle of 45˚ above the horizontal.
The ball hits a tree, a horizontal distance of 122.5m away from the tee box.
How high up the tree did it hit? How fast was is moving when it struck the tree?
(Use a = 10m/s2)
(It’s a tall redwood!)
sy = ?
vfinal = ?
vi = 50m/s 45°
sx = 200m
vix = vi ⋅ cos 45
vix = 35m / s
o
viy = vi ⋅ sin 45o
viy = 35m / s
Ex3) A golf ball is hit at 50m/s at angle of 45˚ above the horizontal.
The ball hits a tree, a horizontal distance of 122.5m away from the tee box.
How high up the tree did it hit? How fast was is moving when it struck the tree?
(Use a = 10m/s2)
(It’s a tall redwood!)
sy = ?
vfinal = ?
vi = 50m/s 45°
sx = 200m
vix = vi ⋅ cos 45
vix = 35m / s
o
viy = vi ⋅ sin 45o
viy = 35m / s
vfy = viy + atup
0 = 35 + (-9.8)tup
tup= 3.5 sec
x2
ttot = 7.0sec
Ex3) A golf ball is hit at 50m/s at angle of 45˚ above the horizontal.
The ball hits a tree, a horizontal distance of 122.5m away from the tee box.
How high up the tree did it hit? How fast was is moving when it struck the tree?
(Use a = 10m/s2)
(It’s a tall redwood!)
sy = ?
vfinal = ?
vi = 50m/s 45°
sx = 200m
vix = vi ⋅ cos 45
vix = 35m / s
o
viy = vi ⋅ sin 45o
viy = 35m / s
vfy = viy + atup
0 = 35 + (-9.8)tup
tup= 3.5 sec
x2
ttot = 7.0sec
sx = vix.ttotal
sx = (35)(7.0)
sx = 245m
Ex3) A golf ball is hit at 50m/s at angle of 45˚ above the horizontal.
The ball hits a tree, a horizontal distance of 122.5m away from the tee box.
How high up the tree did it hit? How fast was is moving when it struck the tree?
(Use a = 10m/s2)
(It’s a tall redwood!)
sy = ?
vfinal = ?
vi = 50m/s 45°
sx = 200m
vix = vi ⋅ cos 45
vix = 35m / s
o
viy = vi ⋅ sin 45o
viy = 35m / s
vfy = viy + atup
0 = 35 + (-9.8)tup
tup= 3.5 sec
x2
ttot = 7.0sec
sx = vix.ttotal
122.5 = (35)(t)
t = 3.5s
sy = ½(vi + vf)t
= ½(35 + 0)3.5
= 61m
sx = vix.ttotal
sx = (35)(7.0) BUT…
sx = 245m
HOW Fast?!?
Ch3 HW#11 p83+ 101,102,107
+ Ch2 Rev WS, side 1
Ch3 HW#11 p83+ 101,102,107
101. A silver dollar thrown downward at angle of 60˚ below the horizontal
from a bridge 50.0m above a river. Initial speed of 40m/s.
How far from the base of the bridge will it strike the water?
102. Someone at a 3rd floor window (12m above ground) hurls a ball downward
at an angle of 45°at a speed of 25m/s. What speed does it hit the ground?
107. A burning firecracker is tossed into the air at an angle of 60˚ at a speed
of 30m/s. How long should the fuse be set to burn if explosion
is 20m away?
107. A burning firecracker is tossed into the air at an angle of 60˚ at a speed
of 30m/s. How long should the fuse be set to burn if explosion
is 20m away?
sx + sy = 20
sy
sx
(vx.t)2 + (viy.t + ½at2)2 = 202
Ch2,3 Rev (red on HW sheet)
4. Corvette from 0 to 26.8 m/s in 4.8 sec. find ave accl.
12. Ball rolls passed a kid with an inst speed of 4.0 m/s.
Ball comes to rest 13.3 sec later. Find ave accl.
Ch2,3 Rev
24. Car goes from rest to 10 km/h in 10 s. At end of 20 s its going 20 km/h.
At the end of 30 s its going 30 km/h. Make a graph of a vs t.
36. Car decelerates from 25 m/s to 15 m/s in 3.5 s. Find ave speed.
58. 2 drivers at rest 100 m apart accl at each other at 2.5 m/s2.
At what time will they collide?
83. A lit firecracker shot straight up at 50 m/s.
Explodes 5 sec later. How high up? How fast?
94. Marble out window at 3m/s. Land in flower pot on fire escape
3 sec later. How far beneath is flower pot?
Ch5 – Graphing ‘s,v,a’ Worksheet
1. Graph of velocity vs time for a car
on a straight road.
a. Inst vel at: t=2:___, t=3.5:___, t=8:___
vel
b. Car’s displacement at:
(m/s)
t=3:___, t=4:___, t=5:___,
t=7:___, t=9:___
c. What is the inst accl at:
t=2:___, t=3.5:___, t=4.5:___,
t=6:___, t=8:___
40
35
30
25
20
15
10
5
1 2 3 4 5 6 7 8 9 10
time (sec)
Ch5 – Graphing Worksheet
1. Graph of velocity vs time for a car
on a straight road.
vel
(m/s)
40
35
30
25
20
15
10
5
1 2 3 4 5 6 7 8 9 10
time (sec)
From a vel-time graph, you can get 3 things:
1. velocity: read it
2. displacement: area
3. acceleration: slope
2) Displacement-time graph
8
a) displacement at:
7
t = 3: ____
t = 7: ____
6
t = 9: ____
s 5
b) Where is vel zero? ____
(m) 4
c) Where is vel max? ____
3
d) vinst @
t = 3: ____
2
t = 6: ____ t = 8: ____
1
t = 10: ____
1 2 3 4 5 6 7 8 9 10
e) Describe motion:
t (s)
t = 0 – 5: ____ t = 5 – 7: ____
t = 7 – 9: ____ t = 9 – 11: ____
3. A football is kicked… 30m/s, 45°, range?
Ch5 – Extra Graphing
1. Graph of velocity vs time for a car
on a straight road.
a. Inst vel at: t=2: __, t=3.5: __, t=8: __
vel
b. Car’s displacement at:
(m/s)
t=3: __, t=4: __, t=5: __,
t=7: __, t=9: ___
c. What is the inst accl at:
t=2: ___, t=3.5: __, t=4.5: ___,
t=6: __, t=8: ___
40
35
30
25
20
15
10
5
1 2 3 4 5 6 7 8 9 10
time (sec)
Ch5 – Graphing Worksheet
1. Graph of velocity vs time for a car
on a straight road.
a. Inst vel at: t=2: 6, t=3.5: 10, t=8: 10
b. Car’s displacement at:
t=3: __, t=4: __, t=5: __,
t=7: __, t=9: ___
c. What is the inst accl at:
t=2: ___, t=3.5: __, t=4.5: ___,
t=6: __, t=8: ___
vel
(m/s)
40
35
30
25
20
15
10
5
1 2 3 4 5 6 7 8 9 10
time (sec)
Ch5 – Graphing Worksheet
1. Graph of velocity vs time for a car
on a straight road.
a. Inst vel at: t=2: 6, t=3.5: 10, t=8: 10
b. Car’s displacement at:
t=3: 15, t=4: 25, t=5: 40,
t=7: 80, t=9: 100
c. What is the inst accl at:
t=2: ___, t=3.5: __, t=4.5: ___,
t=6: __, t=8: ___
vel
(m/s)
40
35
30
25
20
15
10
5
1 2 3 4 5 6 7 8 9 10
time (sec)
Ch5 – Graphing Worksheet
1. Graph of velocity vs time for a car
on a straight road.
a. Inst vel at: t=2: 6, t=3.5: 10, t=8: 10
b. Car’s displacement at:
t=3: 15, t=4: 25, t=5: 40,
t=7: 80, t=9: 100
c. What is the inst accl at:
t=2: +3.3, t=3.5: 0, t=4.5: +10,
t=6: 0, t=8: +10
vel
(m/s)
From a vel-time graph, you can get 3 things:
1. velocity: read it
2. displacement: area
3. acceleration: slope
40
35
30
25
20
15
10
5
1 2 3 4 5 6 7 8 9 10
time (sec)
2. Make a displacement vs time graph for a person
walks 10m in 5s, pauses for 10s to tie a shoe, runs 20m in 5s, pauses
for 10s to tie other shoe, then runs back to start in 10s.
3. Velocity at each time interval:
dist
(m)
40
35
30
25
20
15
10
5
5 10 15 20 25 30 35 40
time (sec)
4. Graph the velocities:
0-5sec: ________
5-10sec: ________
10-15sec: ________
15-20sec: ________
20-25sec: ________
25-30sec: ________
30-35sec: ________
35-40sec: ________
2. Make a displacement vs time graph for a person
walks 10m in 5s, pauses for 10s to tie a shoe, runs 20m in 5s, pauses
for 10s to tie other shoe, then runs back to start in 10s.
3. Velocity at each time interval:
dist
(m)
40
35
30
25
20
15
10
5
0-5sec: _2________
5-10sec: _0________
10-15sec: _0________
15-20sec: _4________
20-25sec: _0________
25-30sec: _0________
30-35sec: _-3________
35-40sec: _-3________
5 10 15 20 25 30 35 40
time (sec)
4. Graph the velocities:
4
3
2
1
vel
0
(m/s)-1
-2
-3
-4
5 10
15 20 25 30 35 40 t(sec)-
dist
(m)
10
9
8
7
6
5
4
3
2
1
1 2 3 4 5 6 7 8 9 10
time (sec)
vel
(m/s)
10
9
8
7
6
5
4
3
2
1
0 1 2 3 4 5 6 7 8 9 10
time (sec)
-1
-2