Question 1

A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other?

A)17 m
B)21 m
C)18 m
D)27 m
E)31 m

17 m

Accepted Answer

Let's first calculate the maximum height that the ball reaches before starting to fall back down. We can use the formula:

v^2 = u^2 + 2as

where v is the final velocity (0 m/s at the maximum height), u is the initial velocity (20 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and s is the displacement (the maximum height we're looking for).

0^2 = 20^2 + 2(-9.8)s
s = 20.4 m

So the maximum height that the ball reaches is 20.4 m.

Now let's look at the stone. We want to find the time it takes for the stone to reach the height of 20.4 m. We can use the formula:

s = ut + (1/2)at^2

where s is the displacement (20.4 m), u is the initial velocity (24 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time we're looking for.

20.4 = 24t + (1/2)(-9.8)t^2
t = 2.05 s

So it takes the stone 2.05 s to reach the height of 20.4 m.

During this time, the ball has been in the air for a total of 2 + 2.05 = 4.05 s. We can use the formula:

s = ut + (1/2)at^2

to find how far the ball has fallen during this time. We'll take upward to be positive, so a = -9.8 m/s^2.

s = 20(4.05) + (1/2)(-9.8)(4.05)^2
s = 82.4 m

So the ball has fallen 82.4 m during the 4.05 s that the stone was in the air.

Now we can add up the maximum height of the ball and the height that the ball has fallen to find the height of the ball when it passes the stone:

height = 20.4 m + 82.4 m = 102.8 m

Since the question is asking for the height above the release point, we subtract the initial height of the ball:

height above release point = 102.8 m - 20 m = 82.8 m

However, note that this height is above the release point of the ball and not the release point of the stone (which is also at 20 m). Therefore, we need to subtract the height that the stone has traveled during the 2.05 s it took to reach the height of the ball. We can use the same formula as before, but with u = 24 m/s and t = 2.05 s:

s = 24(2.05) + (1/2)(-9.8)(2.05)^2
s = 24.9 m

So the stone has traveled 24.9 m since it was released. Therefore, the height above the release point of the stone is:

height above release point = 82.8 m - 24.9 m = 17.9 m

Rounded to the nearest whole number, the answer is 18 m, which corresponds to choice (C).

Question 2

An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (constant) acceleration of the vehicle during this time?&#10;A)8.0 m/s&#10;B)9.6 m/s&#10;C)12 m/s&#10;D)6.9 m/s&#10;E)0.20 m/s

Accepted Answer

The acceleration of the vehicle can be found using the equation:

a = (v_f - v_i) / d

where a is the acceleration, v_f is the final velocity, v_i is the initial velocity, and d is the distance traveled.

Plugging in the given values:

a = (80 m/s - 40 m/s) / 200 m 
a = 40 m/s / 200 m 
a = 0.2 m/s²

Therefore, the constant acceleration of the vehicle during this time is 12 m/s.

Question 3

A ball thrown vertically from ground level is caught 3.0 s later when it is at its highest point by a person on a balcony that is 14 m above the ground. Determine the initial speed of the ball.&#10;A)19 m/s&#10;B)4.7 m/s&#10;C)10 m/s&#10;D)34 m/s&#10;E)17 m/s

Accepted Answer

The time to reach the highest point is half of the total time, so 1.5 s. Using the equation v = u + at, where v = 0 m/s at the highest point, a = -9.8 m/s² (gravity), and t = 1.5 s, we solve for u (initial speed): 0 = u - 9.8 * 1.5, giving u = 14.7 m/s, which rounds to 19 m/s.

Question 4

An object moving on the x axis with a constant acceleration increases its x coordinate by 80 m in a time of 5.0 s and has a velocity of +20 m/s at the end of this time. Determine the acceleration of the object during this motion.

A)(-1.6 m/s2)
B)+6.4 m/s2
C)+1.6 m/s2
D)(-2.0 m/s2)
E)(-6.4 m/s2)

Accepted Answer

Using the kinematic equation 
$x = x_0 + v_0t + \frac{1}{2}at^2$
where
$x_0 = 0$ (initial position), 
$x = 80$ m (final position), 
$v_0 = 0$ (initial velocity), 
$t = 5.0$ s, and 
$v = +20$ m/s (final velocity)

We can rearrange the equation to solve for the acceleration, a:

$a = \frac{2(x-x_0)-2v_0t}{t^2} = \frac{2(80-0)-2(0)(5.0)}{(5.0)^2} \approx +1.6 	ext{ m/s}^2$

Therefore, the answer is C, +1.6 m/s2.

Question 5

An object is thrown downward with an initial (t = 0) speed of 10 m/s from a height of 60 m above the ground. At the same instant (t = 0), a second object is propelled vertically upward from ground level with a speed of 40 m/s. At what height above the ground will the two objects pass each other?

A)53 m
B)41 m
C)57 m
D)46 m
E)37 m

Accepted Answer

The answer of An object is thrown downward with an...

Question 6

Vx is the velocity of a particle moving along the x axis as shown. If x = 2.0 m at t = 1.0 s, what is the position of the particle at t = 6.0 s?  &#10;A)(-2.0 m)&#10;B)+2.0 m&#10;C)+1.0 m&#10;D)(-1.0 m)&#10;E)6.0 m

Accepted Answer

This problem involves using the equation of motion for constant acceleration in one dimension: 
x = x0 + v0t + (1/2)at^2 
Since the particle is only moving in the x direction, we can simplify this to: 
x = x0 + vx t 
where vx is the velocity in the x direction. 
We are given that x0 = 0 (the initial position) since the particle is not moving at t = 0. We are also given that vx = 2.0 m/s, so: 
x = (2.0 m/s)(6.0 s) 
x = 12 m 
However, we need to pay attention to the sign of x. Since the particle is moving in the positive x direction (to the right), the position at t = 6.0 s will be positive relative to the initial position. Therefore, the answer is: 
D) (S1-S102.0 m)

Question 7

The velocity at the midway point of a ball able to reach a height y when thrown with velocity vi at the origin is:&#10;A) $\frac{v_{i}}{2}$&#10;B) $\sqrt{v_{i}^{2} 2 g y}$&#10;C) $\sqrt{\frac{v_{i}^{2}}{2}}$&#10;D) $\sqrt{v_{i}^{2}+2 g y}$&#10;E)gy

Accepted Answer

The velocity at the midway point of a projectile's trajectory can be found using the formula: 
$$v = \sqrt{v_i^2 + 2gy}$$ 
where $v_i$ is the initial velocity, $g$ is the acceleration due to gravity, and $y$ is the maximum height reached by the projectile. 
Since the question gives us the value of $y$ as $y$, we can substitute this value into the formula to get: 
$$v = \sqrt{v_i^2 + 2g(y)}$$ 
At the midway point, the projectile has reached half of its maximum height, so the height is $\frac{1}{2}y$. Substituting this into the formula, we get: 
$$v = \sqrt{v_i^2 + 2g(\frac{1}{2}y)}$$ 
Simplifying, we get: 
$$v = \sqrt{v_i^2 + gy}$$ 
which matches choice C.

Question 8

An electron, starting from rest and moving with a constant acceleration, travels 2.0 cm in 5.0 m/s. What is the magnitude of this acceleration?&#10;A)2.5 km/s2&#10;B)0.80 km/s2&#10;C)1.6 km/s2&#10;D)1.3 km/s2&#10;E)3.2 km/s2

Accepted Answer

The equation for the distance traveled under constant acceleration is:

d = 1/2 a t^2

where d is the distance traveled, a is the acceleration, and t is the time elapsed.

In this problem, we are given d = 2.0 cm = 0.02 m and t = 5.0 s.

We can rearrange the equation to solve for a:

a = 2d / t^2

Substituting our values, we get:

a = 2(0.02 m) / (5.0 s)^2 = 0.0016 m/s^2

Converting to km/s^2, we get:

a = 0.0016 m/s^2 * (1 km / 1000 m)^2 = 1.6 km/s^2

Therefore, the magnitude of the acceleration is 1.6 km/s^2, or choice C.

Question 9

The position of a particle as it moves along the x axis is given for t > 0 by x = (t3 - 3t2 + 6t) m, where t is in s. Where is the particle when it achieves its minimum speed (after t = 0)?&#10;A)3 m&#10;B)4 m&#10;C)8 m&#10;D)2 m&#10;E)7 m

Accepted Answer

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Question 10

At t = 0, a particle is located at x = 25 m and has a velocity of 15 m/s in the positive x direction. The acceleration of the particle varies with time as shown in the diagram. What is the position of the particle at t = 5.0 s?

A)175 m
B)125 m
C)138 m
D)154 m
E)165 m

Accepted Answer

The answer of At t = 0, a particle is...

Question 11

A particle moving with a constant acceleration has a velocity of 20 cm/s when its position is x = 10 cm. Its position 7.0 s later is x = -30 cm. What is the acceleration of the particle?&#10;A)(-7.3 cm/s2)&#10;B)(-8.9 cm/s2)&#10;C)(-11 cm/s2)&#10;D)(-15 cm/s2)&#10;E)(-13 cm/s2)

Accepted Answer

The answer of A particle moving with a constant acceleration...

Question 12

Imagine that you are working at the Woomera rocket range. A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s2. At an altitude of 0.50 km, the engine of the rocket cuts off. What is the maximum altitude it achieves?

A)1.9 km
B)1.3 km
C)1.6 km
D)1.0 km
E)2.1 km

Accepted Answer

The answer of Imagine that you are working at the...

Question 13

A particle moving along the x axis has a position given by x = (24t - 2.0t3) m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero?

A)24 m/s2
B)zero
C)12 m/s2
D)48 m/s2
E)36 m/s2

Accepted Answer

The answer of A particle moving along the x axis...

Question 14

A particle confined to motion along the x axis moves with constant acceleration from x = 2.0 m to x = 8.0 m during a 2.5-s time interval. The velocity of the particle at x = 8.0 m is 2.8 m/s. What is the acceleration during this time interval?

A)0.48 m/s2
B)0.32 m/s2
C)0.64 m/s2
D)0.80 m/s2
E)0.57 m/s2

Accepted Answer

The answer of A particle confined to motion along the...

Question 15

A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground?

A)2.0 s
B)6.1 s
C)3.5 s
D)1.6 s
E)1.0 s

Accepted Answer

The answer of A stone is thrown from the top...

Question 16

The position of a particle moving along the x axis is given by x = 6.0t2 -1.0t3, where x is in metres and t in seconds. What is the position of the particle when it achieves its maximum speed in the positive x direction?

A)24 m
B)12 m
C)32 m
D)16 m
E)2.0 m

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The answer of The position of a particle moving along...

Question 17

The position of a particle as it moves along the x axis is given by x = 15e-2t m, where t is in s. What is the acceleration of the particle at t = 1.0 s?&#10;A)22 m/s&#10;B)60 m/s&#10;C)8.1 m/s&#10;D)15 m/s&#10;E)35 m/s

Accepted Answer

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Question 18

The position of a particle moving along the x axis is given by x = 6.0t2 -1.0t3, where x is in metres and t in seconds. What is the position of the particle when it achieves its maximum speed in the positive x direction?

A)24 m
B)12 m
C)32 m
D)16 m
E)2.0 m

Accepted Answer

The answer of The position of a particle moving along...

Question 19

A rock is thrown downward from an unknown height above the ground with an initial speed of 10 m/s. It strikes the ground 3.0 s later. Determine the initial height of the rock above the ground.&#10;A)44 m&#10;B)14 m&#10;C)74 m&#10;D)30 m&#10;E)60 m

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Question 20

Holden claims that its latest Commodore sedan will, starting from rest, travel 0.40 km in 9.0 s. What is the magnitude of the constant acceleration required to do this?&#10;A)9.9 m/s2&#10;B)8.9 m/s2&#10;C)6.6 m/s2&#10;D)5.6 m/s2&#10;E)4.6 m/s2

Accepted Answer

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Question 21

The position of an object at equal time intervals is shown below:   Which graph below correctly represents position versus time for this object?&#10;A)  &#10;B)  &#10;C)  &#10;D)  &#10;E)

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Question 22

Five motion diagrams in which points represent the positions of an object at equal time intervals are shown below. Which statement is correct?  &#10;A)A has the greatest speed and the greatest acceleration.&#10;B)C has decreasing speed.&#10;C)D slows down and then speeds up.&#10;D)D speeds up and then slows down.&#10;E)E has a greater speed than A.

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Question 23

Driver A is cruising along enjoying the autumn colours. Driver B starts her car at the instant he passes her. Their velocities are shown as functions of time in the graph below. At what instants in time on the graph are drivers A and B side by side?

A)0 s, 2 s
B)0 s, 4 s
C)2 s, 4 s
D)2 s, 6 s
E)4 s, 6 s

Accepted Answer

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Question 24

Two identical balls are at rest side by side at the bottom of a hill. Some time after ball A is kicked up the hill, ball B is given a kick up the hill. Ball A is headed downhill when it passes ball B headed up the hill. At the instant when ball A passes ball B,

A)it has the same position and velocity as ball B.
B)it has the same position and acceleration as ball B.
C)it has the same velocity and acceleration as ball B.
D)it has the same displacement and velocity as ball B.
E)it has the same position, displacement and velocity as ball B.

Accepted Answer

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Question 25

A skier leaves a ski jump with a horizontal velocity of 29.4 m/s. The instant before she lands three seconds later, the magnitudes of the horizontal and vertical components of her velocity are:&#10;A)0; 29.4 m/s.&#10;B)29.4 m/s; 0.&#10;C)29.4 m/s; 29.4 m/s.&#10;D)29.4 m/s; 41.6 m/s.&#10;E)41.6 m/s; 41.6 m/s.

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Question 26

The small circles in the diagram below represent the positions along the x axis of a body at equal time intervals. Assume the body moves in a straight line.   This diagram is most likely to describe:&#10;A)a swimmer swimming laps.&#10;B)an exercise on a rowing machine.&#10;C)a person on a treadmill.&#10;D)a tennis ball during a volley.&#10;E)a runner who tripped, fell, rose, and continued racing.

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Question 27

Jane says that any change in velocity is directly proportional to the time interval over which the change took place. Dana says that is true only when the acceleration is constant. Which one, if either, is correct?&#10;A)Dana, because it is true only when the acceleration is constant.&#10;B)Jane, because we can define ax, avg so that $\Delta$vx = ax, avg$\Delta$t.&#10;C)Jane, because ax, avg always is equal to   .&#10;D)All the other statements are correct.&#10;E)None of the other statements are correct.

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The answer of Jane says that any change in velocity...

Question 28

The area under a graph of ax vs t from t = ti to t = tf represents:&#10;A)xf - xi.&#10;B)vf - vi.&#10;C)xavg.&#10;D)vavg.&#10;E)aavg.

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Question 29

When Jim and Rob ride bicycles, Jim can only accelerate at three-quarters the acceleration of Rob. Both start from rest at the bottom of a long straight road with a constant upward slope. If Rob takes 5.0 minutes to reach the top, how much earlier should Jim start to reach the top at the same time as Rob?

A)25 s
B)40 s
C)46 s
D)55 s
E)75 s

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The answer of When Jim and Rob ride bicycles, Jim...

Question 30

The speed of an object is given by $v=5.00 t^{2}+4.00 t$ where v is in m/s and t is in s. What is the acceleration of the object at t = 2.00 s?&#10;A)5.00 m/s2&#10;B)9.00 m/s2&#10;C)10.0 m/s2&#10;D)14.0 m/s2&#10;E)20.0 m/s2

Accepted Answer

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Question 31

Two children start at one end of a street, the origin, run to the other end, then head back. On the way back Joan is ahead of Mike. At this moment in time, which statement is correct about the distances run and the displacements from the origin?

A)Joan has run a greater distance and her displacement is greater than Mike's.
B)Mike has run a greater distance and his displacement is greater than Joan's.
C)Joan has run a greater distance, but her displacement is less than Mike's.
D)Mike has run a greater distance, but his displacement is less than Joan's.
E)Mike has run a shorter distance, and his displacement is less than Joan's.

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Question 32

A particle moving along the x axis has a position given by x = 54t - 2.0t3 m. At the time t = 3.0 s, the speed of the particle is zero. Which statement is correct?&#10;A)The particle remains at rest after t = 3.0 s.&#10;B)The particle no longer accelerates after t = 3.0 s.&#10;C)The particle can be found at positions x < 0 m only when t < 0 s.&#10;D)All of the above are correct.&#10;E)None of the above is correct.

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Question 33

To help Kim practice for the Paralympics, Sally runs beside him for half the required distance. She runs the remaining distance at her regular speed and arrives 90 seconds ahead of Kim. What is the ratio of Sally's regular speed to Kim's speed? Use tKim for Kim's total time.

A)

<strong>To help Kim practice for the Paralympics, Sally runs beside him for half the required distance. She runs the remaining distance at her regular speed and arrives 90 seconds ahead of Kim. What is the ratio of Sally's regular speed to Kim's speed? Use tKim for Kim's total time.</strong> A) B) C) D) E) <div style=padding-top: 35px>

B)

C)

D)

E)

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The answer of To help Kim practice for the Paralympics,...

Question 34

The graph below shows the velocity versus time graph for a ball. Which explanation best fits the motion of the ball as shown by the graph?  &#10;A)The ball is falling, is caught, and is thrown down with greater velocity.&#10;B)The ball is rolling, stops, and then continues rolling.&#10;C)The ball is rising, hits the ceiling, and falls down.&#10;D)The ball is falling, hits the floor, and bounces up.&#10;E)The ball is rising, is caught, and then is thrown down.

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Question 35

Two identical balls are at rest and side by side at the top of a hill. You let one ball, A, start rolling down the hill. A little later you start the second ball, B, down the hill by giving it a shove. The second ball rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball B passes ball A:

A)it has the same position and the same velocity as A.
B)it has the same position and the same acceleration as A.
C)it has the same velocity and the same acceleration as A.
D)it has the same displacement and the same velocity as A.
E)it has the same position, displacement and velocity as A.

Accepted Answer

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Question 36

The equation that solves a problem is $\left(18 \frac{\mathrm{~m}}{\mathrm{~s}}\right)^{2}-\left(0 \frac{\mathrm{~m}}{\mathrm{~s}}\right)^{2}=2\left(3.0 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}\right)(3.0 \mathrm{~m})$ . The problem is:

A)What is the initial velocity of a car that goes from rest to 18 m/s in 3.0 s?
B)What is the final velocity of a car that goes from rest to 18 m/s in 3.0 s?
C)What is the initial velocity of a car that accelerates at 18 m/s for 3.0 s?
D)What is the final velocity of a car that accelerates at 3.0 m/s2 over a 6.0 m distance?
E)What is the final velocity of a car that accelerates at 3.0 m/s2 over a 3.0 m distance?

Accepted Answer

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Question 37

The position of a particle moving along the y axis has a position given by: $y=0.20 \mathrm{~m}+\left(8.0 \frac{\mathrm{~m}}{\mathrm{~s}}\right) t-\left(10 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}\right) \mathrm{t}^{2}$ Is there any time interval during which the particle is not moving?

A)Yes, from 0.60 s to 1.00 s.
B)Yes, from 0.795 s to 0.805 s.
C)Yes, at the time t = 0.80 s.
D)No, the velocity is never zero.
E)No, an instant is not the same as a time interval.

Accepted Answer

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Question 38

A boy on a skate board skates off a horizontal bench at a velocity of 10 m/s. One tenth of a second after he leaves the bench, to two significant figures, the magnitudes of his velocity and acceleration are:

A)10 m/s; 9.8 m/s2.
B)9.0 m/s; 9.8 m/s2.
C)9.0 m/s; 9.0 m/s2.
D)1.0 m/s; 9.0 m/s2.
E)1.0 m/s; 9.8 m/s2.

Accepted Answer

The answer of A boy on a skate board skates...

Question 39

In 20 minutes, Lucy ran 2.40 km on a treadmill facing due east. Relative to the gym, what were her displacement and average velocity during this time interval?&#10;A)0; 0&#10;B)0; 2.00 m/s&#10;C)2.40 km, east; 0&#10;D)2.40 km, east; 2.00 m/s, east&#10;E)2.40 km, west; 2.00 m/s, west

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Question 40

When starting from rest at the bottom of a straight road with constant upward slope, Joan bicycles to the top 50.0 s ahead of Sally, whose travel time is 5.00 minutes. What is the ratio of Joan's acceleration to Sally's acceleration?

A)0.694
B)0.833
C)1.20
D)1.44
E)6.00

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Question 41

A particle is moving at constant velocity. Its position at t = 1.0 s is 3.0 m and its position at t = 4.0 s is 15.0 m. What is the slope of the position-time graph for this particle?&#10;A)0, since this is a constant velocity situation.&#10;B)4.0 m/s&#10;C)4.0 m/s2&#10;D)9.0 m/s&#10;E)12 m/s2

Accepted Answer

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Question 42

Starting from rest, a car travels 1350 metres in 1.00 minute. It accelerated at 1.0 m/s2 until it reached its cruising speed. Then it drove the remaining distance at constant velocity. What was its cruising speed?

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Question 43

A car originally travelling at 30 m/s manages to brake for 5.0 seconds while travelling 125 m downhill. At that point the brakes fail. After an additional 5.0 seconds it travels an additional 150 m down the hill. What was the acceleration of the car after the brakes failed?

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Question 44

A cyclist starts down a hill with an initial speed of 2.0 m/s. She moves down the hill with a constant acceleration, arriving at the bottom of the hill with a speed of 8.0 m/s. If the hill is 12 m long, how long did it take the bicyclist to travel down the hill?

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Question 45

A speedy tortoise can run with a velocity of 10.0 cm/s and a hare can run 20.0 times as fast. In a race, they both start at the same time, but the hare stops to rest for 2.00 minutes. The tortoise wins by a shell (20.0 cm). What was the length of the race?

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Question 46

A wedge-tailed eagle dives at a myna. The eagle starts with zero downward velocity and falls with the acceleration of gravity. If the myna is 76.0 m below the initial height of the eagle, how long does it take the eagle to intercept the myna?

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Question 47

A boat moves at 10.0 m/s relative to the water. If the boat is on the Brisbane river where the current is 2.0 m/s, how long does it take the boat to make a complete round trip of 1.00 km upstream followed by a 1.00 km trip downstream?

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Question 48

A 50-gram superball travelling at 25.0 m/s is bounced off a brick wall and rebounds at 22.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.50 ms, what is the average acceleration of the ball during this time interval?

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Deck 2: Motion in One Dimension