Motion Along a Line
For constant acceleration the kinematic equations are:
1
Δx = x f − xi = vix Δt + a x Δt 2
2
Δv x = v fx − vix = a x Δt
v fx = vix + 2a x Δx
2
2
Δx = vav , x Δt
Also:
vav , x =
vix + v fx
2
1
Example: If a car traveling at 28 m/s is brought to a full stop
4.0 s after the brakes are applied, find the average
acceleration during braking.
Given: vi = +28 m/s, vf = 0 m/s, and Δt = 4.0 s.
Δv 0 − 28 m/s
aav =
=
= −7.0 m/s 2
Δt
4.0 s
2
Example: A box sliding across a rough surface was found to
have an acceleration of -2.94 m/s2. If the initial speed of the
box is 10.0 m/s, how long does it take for the box to come to
rest?
Know: a= -2.94 m/s2, vix=10.0 m/s, vfx= 0.0 m/s
Want: Δt.
v x = vix + a x Δt = 0
vix
+ 10.0 m/s
Δt = − = −
= 3.40 sec
2
ax
-2.94 m/s
3
Example (text problem 4.8): A train of mass 55,200 kg is
traveling along a straight, level track at 26.8 m/s. Suddenly
the engineer sees a truck stalled on the tracks 184 m ahead.
If the maximum possible deceleration has a magnitude
1.5 m/s2, can the train be stopped in time?
Know: vfx = 0 m/s, vix=26.8 m/s, Δx=184 m
Determine ax and compare to the train’s maximum ax.
v x = vix + 2a x Δx = 0
2
2
− vix
= −1.95 m/s 2
2Δx
2
ax =
4
Visualizing Motion with Constant
Acceleration
Motion diagrams for three carts:
5
Graphs of x, vx, ax for each of the three carts
6
Example (text problem 4.13): A trolley car in New Orleans
starts from rest at the St. Charles Street stop and has a
constant acceleration of 1.20 m/s2 for 12.0 seconds.
(a) Draw a graph of vx versus t.
16
14
12
v (m/sec)
10
8
6
4
2
0
0
2
4
6
8
10
12
14
t (sec)
7
Example continued:
(b) How far has the train traveled at the end of the 12.0
seconds?
The area between the curve and the time axis represents
the distance traveled.
1
Δx = v(t = 12 sec ) ⋅ t
2
1
= (14.4 m/s )(12 s ) = 86.4 m
2
(c) What is the speed of the train at the end of the 12.0 s?
This can be read directly from the graph, vx=14.4 m/s.
8
Free Fall
A stone is dropped from the edge of a cliff; if air resistance
can be ignored, the FBD for the stone is:
y
Apply Newton’s Second Law
x
w
∴a = −g
= −9.8 m/s 2
The stone is in free fall; only the acceleration of gravity
acts on the stone.
9
Example: You throw a ball into the air with speed 15.0 m/s;
how high does the ball rise?
y
viy
Given: viy=+15.0 m/s; ay=-9.8 m/s2
x
ay
To calculate the final height, we
need to know the time of flight.
Time of flight from:
1
Δy = viy Δt + a y Δt 2
2
v fy = viy + a y Δt
10
Example continued:
The ball rises
until vfy= 0.
The height:
v fy = viy + a y Δt = 0
viy
15.0 m/s
Δt = − = −
= 1.53 sec
2
- 9.8 m/s
ay
1
Δy = viy Δt + a y Δt 2
2
(
)
1
2
= (15.0 m/s )(1.53 s ) + − 9.8 m/s 2 (1.53 s )
2
= 11.5 m
11
Example (text problem 4.22): A penny is dropped from the
observation deck of the Empire State Building 369 m above
the ground. With what velocity does it strike the ground?
Ignore air resistance.
y
Given: viy=0 m/s, ay=-9.8 m/s2,
Δy=-369 m
x
ay
Unknown: vyf
369 m
Use:
v fy = viy + 2a y Δy
2
2
= 2a y Δy
v yf = 2a y Δy
12
Example continued:
(
)
v yf = 2a y Δy = 2 − 9.8 m/s 2 (− 369 ) m = 85.0 m/s
(downward)
How long does it take for the penny to strike the ground?
Given: viy=0 m/s, ay=-9.8 m/s2, Δy=-369 m
Unknown: Δt
1
1
2
Δy = viy Δt + a y Δt = a y Δt 2
2
2
2Δy
∴ Δt =
= 8.7 sec
ay
13
Projectile Motion
What is the motion of a struck baseball? Once it leaves
the bat (if air resistance is negligible) only the force of
gravity acts on the baseball.
The baseball has ax = 0 and ay = -g, it moves with
constant velocity along the x-axis and with nonzero,
constant acceleration along the y-axis.
14
Projectile Motion
The motion of an object in a vertical plane under the influence of
gravitational force is known as “projectile motion.”
G
The projectile is launched with an initial velocity v0 .
The horizontal and vertical velocity components are:
v0 x = v0 cos θ 0
g
v0 y = v0 sin θ 0
Projectile motion will be analyzed in a
horizontal and a vertical motion along
the x- and y-axes, respectively. These
two motions are independent of each
other. Motion along the x-axis has
zero acceleration. Motion along the yaxis has uniform acceleration ay = -g.
15
Horizontal Motion: ax = 0
vx = v0 cos θ 0
(eq. 1)
Vertical Motion:
v y = v0 sin θ 0 − gt
ay = − g
(eq. 3)
The velocity along the x-axis does not change:
x = xo + ( v0 cos θ 0 ) t
(eq. 2)
Along the y -axis the projectile is in free fall
gt 2
y = yo + ( v0 sin θ 0 ) t −
2
(eq. 4)
If we eliminate t between equations 3 and 4 we get v − ( v0 sin θ 0 ) = −2 g ( y − yo ) .
2
y
g
2
Here xo and yo are the coordinates
of the launching point. For many
problems the launching point is
taken at the origin. In this case
xo = 0 and yo = 0.
Note: In this analysis of projectile
motion we neglect the effects of
air resistance.
16
(4-8)
The equation of the path:
gt 2
x = ( v0 cos θ 0 ) t (eq. 2)
y = ( v0 sin θ 0 ) t −
2
If we eliminate t between equations 2 and 4 we get:
y = ( tan θ 0 ) x −
g
2 ( v0 cos θ 0 )
2
(eq. 4)
x 2 . This equation describes the path of the motion.
The path equations has the form: y = ax + bx 2 . This is the equation of a parabola.
Note: The equation of the path seems too
complicated to be useful. Appearances can
deceive: Complicated as it is, this equation
can be used as a shortcut in many projectile
motion problems.
(4-9)
17
vx = v0 cos θ 0
(eq. 1)
x = ( v0 cos θ 0 ) t
(eq. 2)
sinϕ
gt 2
O π/2
(eq. 4)
v y = v0 sin θ 0 − gt (eq. 3) y = ( v0 sin θ 0 ) t −
2
Horizontal Range: The distance OA is defined as the horizontal range R
At point A we have: y = 0. From equation 4 we have:
3π/2
gt 2
gt ⎞
⎛
sin
0
sin
−
=
→
−
θ
θ
v
t
t
v
(0
0)
0
⎜ 0
⎟ = 0. This equation has two solutions:
2
2⎠
⎝
Solution 1. t = 0. This solution corresponds to point O and is of no interest.
gt
Solution 2. v0 sin θ 0 − = 0. This solution corresponds to point A.
2
2v0 sin θ 0
From solution 2 we get t =
. If we substitute t in eq. 2 we get
g
2v02
v02
sin θ 0 cos θ 0 = sin 2θ 0 .
R=
g
g
t
A
O
R has its maximum value when θ 0 = 45° :
R
2sin A cos A = sin 2 A
Rmax
v02
=
g
18
ϕ
tA
g
Maximum Height H
v02 sin 2 θ 0
H=
2g
H
The y -component of the projectile velocity is v y = v0 sin θ 0 − gt.
At point A: v y = 0
→ v0 sin θ 0 − gt → t =
H = y (t ) = ( v0 sin θ 0 ) t −
v02 sin 2 θ 0
H=
2g
v0 sin θ 0
g
2
v sin θ 0 g ⎛ v0 sin θ 0 ⎞
gt
= ( v0 sin θ 0 ) 0
− ⎜
⎟ →
2
2⎝
g
g ⎠
2
19
Maximum Height H (encore)
tA
g
H
v02 sin 2 θ 0
H=
2g
We can calculate the maximum height using the third equation of kinematics
2
= 2a ( y − yo ) .
for motion along the y -axis: v y 2 − v yo
In our problem: y0 = 0, y = H , v yo = v0 sin θ 0 , v y = 0 , and a = − g →
v02 sin 2 θ 0
=
−v = −2 gH → H =
.
2g
2g
2
yo
2
v yo
20
Example: An object is projected from the origin. The initial
velocity components are vix = 7.07 m/s, and viy = 7.07 m/s.
Determine the x and y position of the object at 0.2 second
intervals for 1.4 seconds. Also plot the results.
1
1
Δy = viy Δt + a y Δt 2 = a y Δt 2
2
2
Δx = vix Δt
Since the object starts from the origin, Δy and Δx
will represent the location of the object at time Δt.
21
Example continued:
t (sec)
x (meters)
y (meters)
0
0
0
0.2
1.41
1.22
0.4
2.83
2.04
0.6
4.24
2.48
0.8
5.66
2.52
1.0
7.07
2.17
1.2
8.48
1.43
1.4
9.89
0.29
22
Example continued:
This is a plot of the x position (black points) and y position
(red points) of the object as a function of time.
12
10
x,y (m)
8
6
4
2
0
0
0.5
1
1.5
t (sec)
23
Example continued:
This is a plot of the y position versus x position for the
object (its trajectory).
3
2.5
y (m )
2
1.5
1
0.5
0
0
2
4
6
8
10
x (m)
The object’s path is a parabola.
24
Example (text problem 4.36): An arrow is shot into the air
with θ = 60° and vi = 20.0 m/s.
(a) What are vx and vy of the arrow when t=3 sec?
y
The components of the initial
velocity are:
vix = vi cos θ = 10.0 m/s
60°
x
At t = 3 sec:
viy = vi sin θ = 17.3 m/s
v fx = vix + a x Δt = vix = 10.0 m/s
v fy = viy + a y Δt = viy − gΔt = −12.1 m/s
25
Example continued:
(b) What are the x and y components of the displacement of
the arrow during the 3.0 sec interval?
y
r
x
1
Δrx = Δx = vix Δt + a x Δt 2 = vix Δt + 0 = 30.0 m
2
1
1
2
Δry = Δy = viy Δt + a y Δt = viy Δt − gΔt 2 = 7.80 m
2
2
26
Example: How far does the arrow in the previous example
land from where it is released?
The arrow lands when Δy=0.
Solving for Δt:
Δt =
2viy
g
1
Δy = viy Δt − gΔt 2 = 0
2
= 3.53 sec
1
The distance traveled is: Δx = vix Δt + a x Δt 2
2
= vix Δt + 0 = 35.3 m
27