Motion in One Dimension
As you move from one position to another, the length of the straight line drawn from your initial
position to your final position is your displacement.
Displacement is not equal to the distance traveled.
This means that the displacement between two places is the same no matter what path you follow. Any
path connecting the two cities will result in the same displacement, even if you travel around the world.
Displacement
∆x = xf - xi
Displacement = change in position = final position – initial position
Displacement can be positive or negative
Displacement is an example of a quantity that has both magnitude and direction. (a vector)
We can see that ∆x will be positive if xf is larger than xi and negative if xi is larger than xf.
Displacements to the right are considered positive and displacements to the left are considered
negative.
Similarly, upward displacements are positive and downward displacements are considered negative.
Average Velocity
Average velocity is equal to the displacement divided by the time during which the displacement
occurred.
Velocity and speed are not the same in physics. Velocity has both magnitude and direction. Velocity
indicates not only how fast something moves but also the direction of its motion. This means velocity is
a vector.
Average velocity
vave 
x x f  xi

t
t f  ti
Average velocity = change in position/change in time
1
Example 1: Finding Average velocity
Problem:
During a race on level ground, Andra covers 825 meters in 137 s while running due west.
Find Andra’s average velocity.
Solution:
Given:
∆x = 825 meters due west
∆t = 137 s
Unknown:
vave = ?
Equation:
vave 
x 825m

 6.02 m/s due west
t 137 s
Answer
1. Heather and Matthew take 34 minutes to walk eastward along a straight road to a store 2.0 km
away. What is their average velocity in m/s?
Solution:
Given:
Unknown:
Equation:
Answer:
2. Eugene is 75.0 km due south of Salem. If Joe rides from Salem to Eugene on his bike in 6.00 h,
what is his average velocity?
Solution:
Given:
Unknown:
Equation:
Answer:
3. If the bus stop is 0.68 km down the street from the museum and it takes you 9.5 min to walk
north from the bus stop to the museum entrance, what is your average velocity?
Solution:
Given:
2
Unknown:
Equation:
Answer:
4. Simpson drives his car with an average velocity of 24 m/s toward the east. How long will it take
him to drive 560 km on a perfectly straight highway?
Solution:
Given:
Unknown:
Equation:
Answer:
5. How much time would Simpson save by increasing his average velocity to 26 m/s east?
Solution:
Given:
Unknown:
Equation:
Answer:
6. A bus traveled south along a straight path for 3.2 h with an average velocity of 88 km/h, stopped
for 20.0 min, then traveled south for 2.8 h with an average velocity of 75 km/h.
a. What is the average velocity for the total trip?
b. What is the displacement for the total trip?
Solution:
Given:
Unknown:
Equation:
Answer:
3
The quantity that describes how the velocity changes in a given time interval is called acceleration.
Average Acceleration:
a ave 
v v f  vi

t t f  t i
Average acceleration = change in velocity/change in time
Example 2: Average Acceleration
Problem:
As a shuttle bus comes to a normal stop, it slows down from 9.00 m/s to 0.00 m/s in
5.00 s. Find the average acceleration of the bus.
Solution:
Given:
vi = 9.00 m/s
vf = 0.00 m/s
∆t = 5.00 s
Unknown:
aave = ?
Equation:
aave 
v 0.00m / s  9.00m / s  9m / s
= -1.80 m/s2


t
5.00s
5s
1. When the shuttle bus comes to a sudden stop to avoid hitting a dog, it slows from 9.00 m/s to
0.00 m/s in 1.5 s. Find the average acceleration of the bus.
Solution:
Given:
Unknown:
Equation:
Answer:
2. A car travelling initially at 7.0 m/s accelerates to a velocity of 12.0 m/s in 2.0 s. What is the
average acceleration of the car?
Solution:
Given:
4
Unknown:
Equation:
Answer:
3. Turner’s treadmill starts with a velocity of -1.2 m/s and speeds up at regular intervals during a
half-hour workout. After 25 min, the treadmill has a velocity 0f -6.5 m/s. What is the average
acceleration of the treadmill during this period?
Solution:
Given:
Unknown:
Equation:
Answer:
4. If a treadmill starts at a velocity of -2.7 m/s and has a velocity of -1/3 m/s after 5.0 min, what is
the average acceleration of the treadmill?
Solution:
Given:
Unknown:
Equation:
Answer:
5. With an average acceleration of -0.5 m/s2, how long will it take a cyclist to bring a bicycle with
an initial velocity of 13.5 m/s to a complete stop?
Solution:
Given:
Unknown:
Equation:
Answer:
5
The displacement of an object depends on acceleration, initial velocity and time.
The velocity of an object depends on initial velocity, acceleration and time.
Remember that:
v ave 
x
t
For an object moving with constant acceleration, the average velocity is basically the average of the
initial and final velocities:
vave 
v f  vi
2
If we combine these two expressions, we can find an equation for the displacement of an object as a
function of time:
Displacement with constant acceleration:
1
x  (vi  v f )t
2
Displacement = (1/2)(initial velocity + final velocity)(time interval)
Example 3: Displacement with constant acceleration
Problem:
A racing car reaches a speed of 42 m/s. it then begins a uniform negative acceleration,
using its parachute and braking system, and comes to rest 5.5 s later. Find how fr the car
moves while stopping.
Solution:
Given:
vi = 42 m/s
vf = 0 m/s
∆t = 5.5 s
Unknown:
∆x = ?
Equation:
1
1
x  (vi  v f )t  (42m / s  0m / s)(5.5s)
2
2
∆x = 120 m
6
By rearranging the equation for acceleration, we can find an equation for the final velocity of an object.
Remember that: a ave 
v v f  v i

.
t
t
This means that:
Velocity with constant acceleration:
vf = vi + a ∆t
final velocity = initial velocity + (acceleration x time interval)
We can use this information to find another useful expression for the displacement of an object.
Remember that: x 
1
(vi  v f )t
2
Using the expression for vf, we have:
x 
1
(vi  vi  at )t
2
x 
1
(2vi  at )t
2
OR
Displacement with constant acceleration:
1
x  vi t  a(t ) 2
2
Displacement = (initial velocity x time interval) + (1/2) acceleration x (time interval) 2
Example 4: Velocity and displacement with constant acceleration
Problem:
A plane starting at rest at one end of a runway undergoes a constant acceleration of 4.8
m/s2 for 15 s before takeoff. What is its speed at takeoff? How long must the runway be
for the plane to be able to take off?
Solution:
Given:
vi = 0 m/s
a = 4.8 m/s2
Unknown:
vf = ?
∆x = ?
Equation 1:
vf = vi + a ∆t = 0 m/s + 4.8 m/s2 (15 s) = 72 m/s
Equation 2:
1
x  vi t  a(t ) 2 = (0 m/s)(15 s) + (1/2) (4.8 m/s2) (15 s)2
2
∆t = 15 s
∆x = 540 m
7
1. A car accelerated uniformly from rest to a speed of 23. 7 km/h in 6.5 s. Find the distance the car
travels during this time.
Solution:
Given:
Unknown:
Equation:
Answer:
2. When Maggie applies the brakes of her car, the car slows uniformly from 15.00 m/s to 0.00 m/s
in 2.50 s. How many meters before a stop sign must she apply her brakes in order to stop at the
sign?
Solution:
Given:
Unknown:
Equation:
Answer:
3. A jet plane lands with a velocity of 100 m/s and can accelerate at a maximum rate of -5.0 m/s2
as it comes to a rest. Can this plane land at an airport where the runway is 0.8o m long?
Solution:
Given:
Unknown:
Equation:
Answer:
4. A driver in a car traveling at a speed of 78 km/h sees a cat 100 m away on the road. How long
will it take for the car to accelerate constantly to a stop in exactly 99 m?
Solution:
Given:
Unknown:
8
Equation:
Answer:
5. A car enters the freeway with a speed of 23 km/h and accelerates to a speed of 86 km/h in 3.5
min. How far does the car move while accelerating?
Solution:
Given:
Unknown:
Equation:
Answer:
6. A car with initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s2 for 3.6 s. Find the
final speed and the displacement of the car during this time.
Solution:
Given:
Unknown:
Equation:
Answer:
7. An automobile with an initial speed of 4.30 m/s accelerates at the rate of 3.00 m/s2. Find the
final speed and the displacement after 5.0 s.
Solution:
Given:
Unknown:
Equation:
Answer:
8. A car starts at rest and travels for 5.0 s with a uniform acceleration of -1.5 m/s2. What is the final
velocity of the car? How far does the car travel in this time interval?
Solution:
Given:
9
Unknown:
Equation:
Answer:
9. A driver of a car traveling at -15 m/s applies the brakes, causing a uniform acceleration of 2.0
m/s2. If the brakes are applied for 2.5 s, what is the velocity of the car at the end of the braking
period? How far has the car moved during the braking period?
Solution:
Given:
Unknown:
Equation:
Answer:
10