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1D Kinematics #3 (7992916)
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Question Details
OSColPhys1 2.4.016. [3203536]
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OSColPhys1 2.5.020. [3356035]
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A cheetah can accelerate from rest to a speed of 29.0 m/s in 4.00 s. What is its acceleration?
7.25 m/s2
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2.
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An Olympic-class sprinter starts a race with an acceleration of 4.10 m/s2.
(a) What is her speed 2.00 s later?
8.2 m/s
(b) Sketch a graph of her position vs. time for this period.
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3.
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OSColPhys1 2.5.028. [2153674]
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OSColPhys1 2.5.033. [2153844]
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A powerful motorcycle can accelerate from rest to 27.8 m/s (62 mi/h) in only 2.90 s.
(a) What is its (constant) acceleration?
9.59 m/s2
(b) How far does it travel in that time?
40.3 m
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4.
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An unwary football player collides with a padded goalpost while running at a velocity of 8.50 m/s and comes to a full stop
after compressing the padding and his body 0.150 m.
(a) How long does the collision last?
0.0353 s
(b) What is his deceleration?
241 m/s2
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5.
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OSColPhys1 2.7.041. [3346547]
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Assume air resistance is negligible unless otherwise stated.
Calculate the displacement and velocity at the following times for a ball thrown straight up with an initial velocity of
12.2 m/s. Take the point of release to be y0 = 0.
(a)
0.500 s
y1 =
4.88 m
v1 =
7.3 m/s
(b)
1.00 s
y2 =
7.3 m
v2 =
2.4 m/s
(c)
1.50 s
y3 =
7.28 m
v3 =
-2.5 m/s
(d)
2.00 s
y4 =
4.8 m
v4 =
-7.4 m/s
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6.
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OSColPhys1 2.7.046. [2153275]
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A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s,
and her takeoff point is 1.20 m above the pool.
(a) How long are her feet in the air?
0.888 s
(b) What is her highest point above the board?
0.459 m
(c) What is her velocity when her feet hit the water?
-5.7 m/s
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7.
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OSColPhys1 2.P.016.Tutorial.WA. [2707319]
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A remote controlled toy car starts from rest and begins to accelerate in a straight line. The figure below represents
"snapshots" of the car's position at equal 0.5 s time intervals. (Assume the positive direction is to the right. Indicate the
direction with the sign of your answer.)
(a) What is the car's average velocity in the interval between t = 1.0 s to t = 1.5 s?
1 m/s
(b) Using data from t = 1.0 s to t = 2.0 s, what is the car's acceleration at t = 1.5 s?
0.8 m/s2
(c) Is the car's speed increasing or decreasing with time?
increasing
decreasing
constant
not enough information
Tutorial
Solution or Explanation
(a) Apply the definition of average velocity, change in position divided by change in time, to the interval from 1.0 s to
1.5 s.
v=
Δx = xf − xi = 0.9 m − 0.4 m = 1.0 m/s
tf − ti
1.5 s − 1.0 s
Δt
(b) Apply the definition of average acceleration, change in velocity divided by change in time, to the given interval. This
requires an initial and final velocity. The initial velocity is the velocity in the interval from 1.0 s to 1.5 s, found in part (a).
The final velocity in the interval from 1.5 s to 2.0 s can be found from:
v=
Δx = xf − xi = 1.6 m/s − 0.9 m/s = 1.4 m/s.
tf − ti
2.0 s − 1.5 s
Δt
The average acceleration is then given by a = Δv/Δt. The time interval Δt is measured from the midpoint of the first
interval to the midpoint of the second, and so is therefore 0.5 s.
a=
Δv = vf − vi = 1.4 m/s − 1.0 m/s = 0.8 m/s2
0.5 s
Δt
Δt
(c) Since the velocity and acceleration are both in the same direction (the positive direction in this case), the car's speed
(or magnitude of velocity) is increasing.
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OSColPhys1 2.P.019.WA. [2707267]
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Starting from rest, a runner at a track meet reaches a speed of 8.2 m/s in 1.9 s. How far does she run during this time,
assuming her acceleration is uniform?
7.79 m
Solution or Explanation
Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the
unrounded values.
We know the initial speed is 0, and we are given the final speed and the time. Using this we can find the runner's
acceleration, which we could then use to find the distance. From the definition of acceleration,
a=
Δv = v − v0 = 8.2 m/s − 0 = 4.32 m/s2.
1.9 s
Δt
Δt
Since we now know the acceleration, time, and initial speed, and we are assuming constant acceleration, we can use the
relation:
x = x0 + v0Δt +
1
1
aΔt2 = 0 + 0 + (4.32 m/s2)(1.9 s)2 = 7.79 m.
2
2
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9.
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OSColPhys1 2.P.023.WA. [2707309]
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An aircraft, traveling northward, lands on a runway with a speed of 65 m/s. Once it touches down, it slows to 5.9 m/s over
720 m of runway. What is the average acceleration (magnitude and direction) of the plane during landing? Take the
positive direction to be northward. (Indicate the direction with the sign of your answer.)
-2.91 m/s2
Solution or Explanation
We are given the initial speed, final speed, and displacement. Assuming a constant acceleration, we can use the following
equation to solve for the acceleration:
v2 = v02 + 2aΔx
2aΔx = v2 − v02
a =
v2 − v02
2Δx
(5.9 m/s)2 − (65 m/s)2
a =
2(720 m)
a = −2.91 m/s2
The sign of acceleration is negative: the plane moves north, but is slowing down, so the acceleration is in the opposite
direction.
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10.
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Question Details
OSColPhys1 2.P.027.WA. [2707335]
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Starting from rest, a truck travels in a straight line for 9.0 s with a uniform acceleration of +1.8 m/s2. The driver then
applies the brakes for 2.0 s, causing a uniform acceleration of −2.0 m/s2 over that time.
(a) What is the truck's speed at the end of the braking period?
12.2 m/s
(b) What is the total distance traveled by the truck (from the point where it started at rest to the end of the
braking period)?
101 m
Solution or Explanation
(a) First, we will determine the velocity of the truck after the initial period of acceleration and before braking. Once we
have this, we can use it as the initial velocity during the period of braking. Initially the truck is at rest (v0 = 0). We
know the initial acceleration and time interval, so we can find the final velocity using:
v = v0 + aΔt = 0 + (1.8 m/s2)(9.0 s) = 16.2 m/s.
This is now the initial velocity for the next phase, where again we know the acceleration and time, and so can find the
final velocity:
v = v0 + aΔt = 16.2 m/s + (−2.0 m/s2)(2.0 s) = 12.2 m/s.
(b) Again we can analyze the motion in two stages by finding the distance traveled over each period of constant
acceleration, using one of the constant acceleration relations. For the first period:
x = x0 + v0Δt +
x=0+0+
1
aΔt2
2
1
(1.8 m/s2)(9.0 s)2 = 72.9 m.
2
For the second period of the motion, we have the initial velocity, found in the first step of our analysis for part (a), the
time, and the acceleration. We will simply take the initial position to be the distance we just found. Then the answer will
be the total distance.
x = x0 + v0Δt +
1
aΔt2
2
x = 72.9 m + (16.2 m/s)(2.0 s) +
1
(−2.0 m/s2)(2.0 s)2 = 101 m
2
(Note that the displacement Δx of the second phase is 28.4 m. Adding it to the first displacement again gives 101 m.)
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11.
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OSColPhys1 2.P.033.WA. [2707265]
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You launch a model rocket from ground level. It moves directly upward with a constant acceleration of 81.0 m/s2 for 1.50
seconds, at which point it runs out of fuel. Assuming air resistance on the rocket is negligible, what is the maximum
altitude (above the ground) achieved by the rocket?
844 m
Solution or Explanation
Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the
unrounded values.
First, we can find the distance traveled while the rocket engine is on. We know the acceleration, initial speed (it starts
from rest) and time, and can use one of the constant acceleration relations:
y = y0 + v0Δt +
1
aΔt2 = 0 + 0 +
2
1
(81.0 m/s2)(1.50 s)2 = 91.1 m.
2
From this point on, the engine is off, and the rocket is in free fall. We could find the distance traveled while it is in free
fall, but we first need to find the velocity when the engine goes out. We can use:
v = v0 + aΔt = 0 + 81.0 m/s2(1.50 s) = 122 m/s.
This final velocity for the first phase of the flight is now the initial velocity for the second. The rocket will continue to go
up for a while, but because the acceleration is now downward, the speed will decrease until it reaches zero momentarily
and reverses direction. The velocity at the maximum height is therefore zero. Knowing the initial velocity, final velocity,
and acceleration, we can find the displacement for the second phase of the flight, using the relationship:
v2 = v02 + 2aΔy.
Solving for displacement:
2
2
2
Δy = v − v0 = 0 − (122 m/s) = 753.2 m.
2a
2(−9.8 m/s2)
This is, however, the displacement from the point where the engine turns off. To find the total height, add this to the
height we found at the end of the first phase:
91.1 m + 753.2 m = 844 m.
Supporting Materials
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12.
Question Details
OSColPhys1 2.P.042.Tutorial.WA. [2707433]
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(a) The graph below plots the position versus time for an object moving in one dimension along the x direction.
What is the speed (magnitude of velocity) of the object at t = 2.5 s?
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1.33 m/s
What direction is the object moving along that time?
+x
−x
The object is not moving.
What is the acceleration of the object at t = 2.5 s? (Indicate the direction with the sign of your answer.)
0 m/s2
Is the speed increasing, decreasing or constant at that time?
increasing
decreasing
constant
(b) The graph below plots the velocity versus time for an object, different from the one in part (a), moving in one
dimension along the x direction.
What is the speed (magnitude of velocity) of the object at t = 2.5 s?
6 m/s
What direction is the object moving along that time interval?
+x
−x
The object is not moving.
What is the acceleration of the object at t = 2.5 s? (Indicate the direction with the sign of your answer.)
-2.67 m/s2
Is the speed increasing, decreasing or constant at that time?
increasing
decreasing
constant
Tutorial
Solution or Explanation
(a) The slope of the position graph is constant from 1 s to 4 s, indicating a constant velocity. This means that the
instantaneous velocity at 2.5 s is the same as the average velocity from 1 s to 4 s. Average velocity is:
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v=
Δx = xf − xi = 1 m − 5 m = −1.33 m/s.
tf − ti
4s−1s
Δt
The speed is the magnitude, or 1.33 m/s. The sign of the velocity indicates that the direction is negative x. The constant
slope indicates a constant velocity; the acceleration is zero, and the speed is constant.
(b) Since this is a graph of velocity vs. time, the velocity can simply be read from the graph. At 2.5 s the velocity is
approximately 6.0 m/s. The speed is therefore also 6.0 m/s, and the velocity is in the positive x direction.
Although the object moves in the positive x direction for the entire time, the speed decreases from 1 s to 4 s at a constant
rate, as is evident from the constant slope. The acceleration at 2.5 s can, therefore, be found from the average
acceleration from 1 s to 4 s:
a=
Δv = vf − vi = 2 m/s − 10 m/s = −2.67 m/s2.
tf − ti
4s−1s
Δt
The velocity is in the positive direction, but the acceleration is in the opposite direction, which means the speed
decreases.
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13.
Question Details
OSColPhys1 2.P.045.WA. [2707296]
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A particle is restricted to move along one dimension, the x-axis. The graph below plots the velocity of the particle as a
function of time.
(a) What is the acceleration of the particle during the following intervals? Indicate the direction with the sign of
your answer.
0 s to 2 s
1.5 m/s2
2 s to 4 s
0 m/s2
4 s to 5 s
3 m/s2
5 s to 9 s
-2 m/s2
9 s to 10 s
2 m/s2
10 s to 12 s
0 m/s2
(b) At the following instants in time, what is the direction of the particle's velocity?
t = 1.0 s
---Select---
positive x
t = 3.0 s
---Select---
positive x
t = 6.0 s
---Select---
positive x
t = 8.0 s
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---Select---
The magnitude is zero.
t = 8.5 s
---Select---
negative x
t = 9.5 s
---Select---
negative x
t = 11.0 s
---Select---
The magnitude is zero.
Solution or Explanation
(a) Average acceleration is defined as a = Δv/Δt, the change in velocity over change in time. For each segment of the
graph, calculate the change in v, the vertical axis, divided by the change in t, the horizontal axis. In other words, find the
slope of each segment. The sign of the slope indicates the direction of the acceleration.
0 s to 2 s:
a =
2 s to 4 s:
a =
4 s to 5 s:
a =
5 s to 9 s:
a =
9 s to 10 s:
a =
10 s to 12 s:
a =
3 m/s − 0 m/s
= 1.5 m/s2
2s−0s
3 m/s − 3 m/s
= 0 m/s2
4s−2s
6 m/s − 3 m/s
= 3.0 m/s2
5s−4s
−2 m/s − 6 m/s
= −2.0 m/s2
9s−5s
0 m/s − (−2 m/s)
= 2.0 m/s2
10 s − 9 s
0 m/s − 0 m/s
= 0 m/s2
12 s − 10 s
(b) Since this is a plot of velocity vs. time, just read the direction off of the sign of the value of the graph, not the slope
of the graph. (The slope gives the acceleration, not the velocity.) While the velocity magnitude (or speed) may be
increasing, decreasing or constant between 0 s and 8 s, it is always in the positive direction. At 8 s it is zero, but only
momentarily, as it then reverses to the negative direction. From 8 s to 9 s the speed increases in the negative direction,
then from 9 s to 10 s it decreases, but is still in the negative direction. From 10 s onward, the velocity is a constant zero
(motionless).
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Name (AID): 1D Kinematics #3 (7992916)
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