This Week
Errors Lecture (in the lab)
Next Week
Tutorial and Test 1
(chapters 1 and 2)
Questions for tutorial 1
Cutnell edition 8
Ch 1: 6
Ch 1: 53
Ch 2: 10
Ch 2: 54
Cutnell edition 7
Ch 1: 54
Ch 1: 47
Ch 2: 8
Ch 2: 47
Wednesday, September 16, 2009
1
Chapter 2: Kinematics in One Dimension
Will cover motion in a straight line with constant acceleration:
• Displacement – not always the same as distance travelled
• Speed, velocity, acceleration
• Equations of motion in one dimension
• Free fall under gravity - which way is up?
• Graphical representation
Wednesday, September 16, 2009
2
Displacement, average speed, velocity
Car starts at xo at time to,
reaches x at time t
Distance travelled = x - xo
Average speed =
x − x0
Distance
=
Elapsed time t − t0
Average velocity =
Displacement, !�x =�x −�x0
!�x
Displacement
=
Elapsed time t − t0
Wednesday, September 16, 2009
3
Displacement and distance not necessarily
the same
Example: Car travels 50 km to east, then 20 km to west in 1 hour.
Distance travelled = 50 + 20 = 70 km
Average speed = 70 km/h
30 km
20 km
50 km
Displacement = 30 km to the east
Average velocity = 30 km/h to east
Wednesday, September 16, 2009
4
Example: A car makes a trip due north for 3/4 of the time and due south for
1/4 of the time. The average northward velocity has a magnitude of 27 m/s.
The average southward velocity has a magnitude of 17 m/s.
What is the average velocity for the entire trip?
Put T = time for the entire trip.
x1 = (3T/4) x (27 m/s)
x2 = (T/4) x (17 m/s)
Average velocity = Displacement/Time
= (x1 – x2)/T, to the north
Net
= 3 ! 27/4 – 17/4 m/s
displacement
= 16 m/s, to the north
Wednesday, September 16, 2009
5
Clicker Questions: Focus on Concepts, Question 2
Three runners start at the same place. Shaun runs 4.0 km due east and then
runs 1.0 km due west. Mark runs 3.0 km due east. Jeff runs 2.0 km due west
and then runs 5.0 km due east. Which of the following is true concerning the
displacement of each runner?
A) Shaun’s displacement equals Mark’s displacement, but Jeff’s displacement is
different.
B) Shaun, Mark, and Jeff have the same displacements.
C) Shaun’s displacement equals Jeff’s displacement, but Mark’s displacement is
different.
D) Shaun, Mark, and Jeff have different displacements.
E) Mark’s displacement equals Jeff’s displacement, but Shaun’s displacement is
different.
Answer B)
Wednesday, September 16, 2009
6
Clicker Question: Focus on Concepts, Question 4
A cyclist races around a circular track and covers the same number of
metres per second everywhere. Which one of the following is true?
A) Neither the speed nor the velocity of the cyclist is constant.
B) The velocity of the cyclist is constant.
C) The speed of the cyclist is constant.
D) The speed and the velocity of the cyclist are constant.
Hint: velocity has a magnitude (speed) and a direction
Answer C)
Wednesday, September 16, 2009
7
Clicker Question: Focus on Concepts, Question 5
A boat moves 3.0 km due north and then moves 1.0 km due south. The time for
this trip is half an hour. What is the average velocity of the boat?
A) 8.0 km/h
B) 2.0 km/h, due south
C) 4.0 km/h, due north
D) 4.0 km/h
E) 8.0 km/h, due north
Answer C)
Wednesday, September 16, 2009
8
Instantaneous Velocity
The velocity measured during a vanishingly small time
interval. That is, the velocity at a particular instant in time.
�v = lim
!t→0
!�x
!t
This differs from the average velocity because the average
is measured over an extended time during which the object
may be changing velocity.
Wednesday, September 16, 2009
9
Acceleration
Average acceleration =
Change in velocity �v −�v0
=
Elapsed time
t − t0
Instantaneous acceleration = lim
!t→0
!�v
!t
Any change of velocity, including slowing down, is an “acceleration”.
Wednesday, September 16, 2009
10
Average acceleration
9−0
= 9 km/h per second
1−0
9000
= 2.5 m/s2
=
3600
ā =
ā =
18 − 9
= 9 km/h per second
2−1
Wednesday, September 16, 2009
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Clicker Question: Focus on Concepts, Question 6
The velocity of a train is 80.0 km/h, due west. One and a half hours later its
velocity is 65.0 km/h, due west. What is the train’s average acceleration?
A) 43.3 km/h2, due east
B) 53.3 km/h2, due east
C) 10.0 km/h2, due east
D) 10.0 km/h2, due west
E) 43.3 km/h2, due west
Wednesday, September 16, 2009
Answer C)
12
Clicker Question
Two cars are moving in a straight section of a highway. The
acceleration of the first car is greater than the acceleration of the
second car and both accelerations have the same direction.
Which one of the following is true?
a) The velocity of the first car is always greater than the velocity of
the second car.
b) The velocity of the second car is always greater than the velocity
of the first car.
c) In the same time interval, the velocity of the first car changes by a
greater amount than the velocity of the second car.
d) In the same time interval, the velocity of the second car changes by
a greater amount than the velocity of the first car.
Answer C)
Wednesday, September 16, 2009
13
Equations of Motion
Consider an object that has speed v0 at time t = 0.
It is accelerated in a straight line at a constant rate to speed v at
time t.
Acceleration:
a=
v − v0
, so v = v0 + at
t
(1)
Average speed:
v̄ =
v + v0 v0 + at + v0
x − x0
=
=
2
2
t
1
x − x0 = v0t + at 2
2
Wednesday, September 16, 2009
(2)
14
From previous page:
v̄ =
x − x0 v + v0 v0 + at + v0
=
=
t
2
2
1
x − x0 = (v + v0)t
2
Multiply –
And:
(1)
(3)
(3)
v − v0 = at
v + v0 x − x0
=
2
t
(v − v0) ×
(x − x0)
(v + v0)
= at ×
2
t
v2 − v20 = 2a(x − x0)
(4)
Wednesday, September 16, 2009
15
The famous four formulae
v = v0 + at
(1)
1
x − x0 = v0t + at 2
2
(2)
1
x − x0 = (v + v0)t
2
(3)
v2 − v20 = 2a(x − x0)
(4)
You will definitely need to know these!
Wednesday, September 16, 2009
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Example: A runner accelerates to a velocity of 5.36 m/s due west in 3
seconds. His average acceleration is 0.640 m/s2, also directed due west.
What was his velocity when he began accelerating?
Take quantities pointing to the east (right) as positive.
v0 = ?
v = −5.36 m/s
a = −0.640 m/s2
t=3s
(1)
v − v0 = at
So:
v0 = v − at = −5.36 − (−0.640) × 3 = −3.44 m/s
Answer: 3.44 m/s due west.
Wednesday, September 16, 2009
17
Clicker Question: Focus on Concepts, Question 24
The graph accompanying this problem shows a three-part motion. For each of
the three parts, A, B, and C, identify the direction of the motion. A positive
velocity denotes motion to the right.
A) A left, B right, C right
B) A right, B left, C right
C) A right, B right, C left
D) A right, B left, C left
E) A left, B right, C left
Wednesday, September 16, 2009
Answer B)
18
Clicker Question
A runner runs half the remaining distance to the finish line every ten
seconds. She runs in a straight line and does not ever reverse her
direction.
Does her acceleration have a constant magnitude?
Hint
• Suppose she starts at 2L from finish, covers a distance L in the first 10 s.
Average speed is v1 = L/10.
• In the second 10 s, she covers a distance L/2.
Average speed is v2 = L/20, and so on..
Average acceleration is (change in speed)/time
–!is it same from v1 to v2 as it is from v2 to v3
in the next 10 s ?
a) the acceleration
is constant
b) the acceleration
increases
c) the acceleration
decreases
Answer C)
Wednesday, September 16, 2009
19
Example: A car accelerates from rest to a final speed in two stages.
Each stage takes the same time T.
In stage 1, the car’s acceleration is a = 3.0 m/s2 and ends at speed v1.
At the end of stage 2, the car is travelling 2.5 times as fast as at the
end of stage 1. The acceleration is a .
Question: what is the acceleration during stage 2?
v0
a = 3 m/s2
T
1
v1
a’ = ?
T
v2 = 2.5v1
2
Hints:
• What is v1 in terms of a and T ?
• What is the final speed in terms of v1, a and T ?
Wednesday, September 16, 2009
20