Question 1

To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of the pole after 0.5 s and then reaches the top of the pole again after a total elapsed time of 4.1 s. How high is the pole above the point where the ball was launched?
A)10 m
B)13 m
C)16 m
D)18 m
E)26 m

Accepted Answer

The answer of To determine the height of a flagpole,...

Question 2

FIGURE 2-8 
-Fig. 2-8 shows the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the total distance run by the player in the 10 s shown in the graph.
A)20 m
B)18 m
C)16 m
D)14 m
E)12 m

Accepted Answer

The total distance run by the player is the area under the velocity-versus-time graph. The area of the triangle is (1/2) * base * height = (1/2) * 10 s * 2 m/s = 10 m. The area of the rectangle is 10 s * 2 m/s = 20 m. The total distance run by the player is 10 m + 20 m = 30 m.

Question 3

Two objects are dropped from a bridge, an interval of 1.00 s apart. What is their separation 1.00 s after the second object is released?
A)4.90 m
B)7.35 m
C)9.80 m
D)14.7 m
E)19.8 m

Accepted Answer

The answer of Two objects are dropped from a bridge,...

Question 4

In a relay race, runner A is carrying the baton and has a speed of 3.4 m/s. When he is 25 m behind the starting line, runner B starts from rest and accelerates at 0.140 m/s². How fast is B traveling when A overtakes her? A)0.10 m/s B)0.33 m/s C)1.3 m/s D)2.0 m/s E)A never catches up.

Accepted Answer

Runner A travels at a constant speed, while runner B accelerates from rest. To find when A overtakes B, set their distances equal: 3.4t = 0.5(0.140)t^2 + 25. Solving for t gives the time when A overtakes B. Then, use v = at to find B's speed at that time.

Question 5

An object is thrown upwards with a speed of 14 m/s. How long does it take to reach a height of 5.0 m above the projection point while descending?
A)0.42 s
B)1.2 s
C)2.4 s
D)3.1 s
E)4.2 s

Accepted Answer

The answer of An object is thrown upwards with a...

Question 6

FIGURE 2-8 
-Fig. 2-8 shows the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the net displacement of the player for the 10 s shown on the graph.
A)20 m
B)18 m
C)16 m
D)14 m
E)12 m

Accepted Answer

The answer of FIGURE 2-8 
-Fig. 2-8 shows the velocity-versus-time...

Question 7

A rock is dropped from a vertical cliff. The rock takes 7.00 s to reach the ground below the cliff. What is the height of the cliff?
A)80.1 m
B)240 m
C)100 m
D)26.2 m
E)481 m

Accepted Answer

The answer of A rock is dropped from a vertical...

Question 8

The acceleration of an object as a function of time is given by a(t) = (3.00 m/s³)t. If the object has a velocity 1.00 m/s at time t = 1.00 s, what is the displacement of the object between time t = 2.00 s and time t = 4.00 s? A)33.0 m B)30.0 m C)36.0 m D)27.0 m E)Not enough information is given to determine the displacement of the object between these two times.

Accepted Answer

The answer of The acceleration of an object as a...

Question 9

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the car a deceleration of 3.50 m/s². If the car comes to a stop in a distance of 30.0 m, what was the car's original speed? A)10.2 m/s B)14.5 m/s C)105 m/s D)210 m/s E)315 m/s

Accepted Answer

We can use the kinematic equation vf^2 = vi^2 + 2ad, where vf is the final velocity (0 m/s), vi is the initial velocity (what we want to find), a is the acceleration (-3.50 m/s^2), and d is the distance (30.0 m). Plugging in the values and solving for vi, we get vi = 14.5 m/s. Therefore, the car's original speed was 14.5 m/s.

Question 10

The acceleration of an object as a function of time is given by a(t) = (1.00 m/s²)t². If displacement of the object between time t = 1.00 s and time t = 2.00 s is 15.0 m, what is the velocity of the object at time t = 0.00 s? A)1.25 m/s B)3.75 m/s C)6.25 m/s D)13.5 m/s E)Not enough information is given to determine the velocity of the object at time t = 0.00 s.

Accepted Answer

The answer of The acceleration of an object as a...

Question 11

The acceleration of an object as a function of time is given by a(t) = (3.00 m/s³)t. If the object is at rest at time t = 0.00 s, what is the velocity of the object at time t = 5.00 s? A)15.0 m/s B)37.5 m/s C)0.00 m/s D)12.0 m/s E)75.0 m/s

Accepted Answer

The velocity is the integral of acceleration with respect to time. Integrating a(t) = (3.00 m/s³)t from 0 to 5 s gives v(t) = (3.00/2)t². Evaluating this at t = 5 s gives v(5) = (3.00/2)(5)² = 37.5 m/s.

Question 12

If an object was freely falling, from what height would it need to be dropped to reach a speed of 70.0 m/s before reaching the ground?
A)250 m
B)322 m
C)189 m
D)500 m
E)712 m

Accepted Answer

The answer of If an object was freely falling, from...

Question 13

A rock is dropped from a vertical cliff. The rock takes 3.00 s to reach the ground below the cliff. A second rock is thrown vertically from the cliff. It takes 2.00 s to reach the ground below the cliff from the time it is released. With what velocity was the second rock released?
A)4.76 m/s upward
B)5.51 m/s downward
C)12.3 m/s upward
D)4.76 m/s downward
E)12.3 m/s downward

Accepted Answer

The time taken by the second rock is less than the first, indicating it was thrown downward. Using the equation of motion $ s = ut + \frac{1}{2}gt^2 $, and knowing the first rock's time gives the height, solve for the initial velocity of the second rock.

Question 14

A ball is dropped from somewhere above a window that is 2.00 m in height. As it falls, it is visible to a person looking through the window for 200 ms as it passes by the 2.00 m height of the window. From what height above the top of the window was the ball dropped?
A)8.32 m
B)1.87 m
C)4.15 m
D)4.76 m
E)6.78 m

Accepted Answer

The ball is visible for 200 ms while passing the 2.00 m window. Using the equations of motion, we can calculate the initial height from which the ball was dropped. The correct calculation shows that the ball was dropped from a height of 4.15 m above the top of the window.

Question 15

In a relay race, runner A is carrying the baton and has a speed of 3.00 m/s. When he is 25.0 m behind the starting line, runner B starts from rest and accelerates at 0.100 m/s². How long afterwards will A catch up with B to pass the baton to B? A)5.17 s B)10.0 s C)11.9 s D)20.4 s E)A never catches up.

Accepted Answer

The answer of In a relay race, runner A is...

Question 16

Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.50 s and reaches the level of the top of the pole after a total elapsed time of 4.10 s. What was the speed of the ball at as it passed the top of the flagpole?
A)6.40 m/s
B)16.2 m/s
C)17.6 m/s
D)29.0 m/s
E)33 m/s

Accepted Answer

Let's first find the height of the flagpole. 
During the first half of the trip, when the ball is going up, we can use the equation:
$h = v_i t + \frac{1}{2}at^2$
Where $v_i$ is the initial velocity (which we assume to be positive as the ball goes up), $t$ is the time it takes to reach the top of the flagpole (0.5 s), $a$ is the acceleration due to gravity (-9.8 m/s^2) and $h$ is the height of the flagpole.
Plugging in the values, we get:
$h = (v_i)(0.5) + \frac{1}{2}(-9.8)(0.5)^2$
$h = 1.225v_i - 1.225$
During the second half of the trip, when the ball is coming back down, we can use the same equation with the same values of $t$ and $a$, but with a different initial velocity ($v_f$, which we assume to be negative as the ball is coming back down). We also need to add the height of the flagpole (h) to get the total height the ball travels:
$h = v_ft+\frac{1}{2}at^2 + h$
$s \ (total \ height) = h + h = 2h = v_f(4.1-0.5) + \frac{1}{2}(-9.8)(3.6)^2 + 1.225v_i - 1.225$
$s=3.675v_f+45.36+1.225v_i - 1.225$
$s=3.675v_f+44.135v_i+44.135$
We know that the total height is twice the height of the flagpole. So, we can set $s$ equal to $2h$ and solve for $v_i$:
$2h = 3.675v_f+44.135v_i+44.135$
$h = 1.8375v_f+22.0675v_i+22.0675$
$1.225v_i - 1.225 = 1.8375v_f+22.0675v_i+22.0675$
$-0.6125v_i = 1.8375v_f + 1.225$
$v_i = -3v_f - 2$
Now that we know $v_i$ in terms of $v_f$, we can use the fact that the velocity is 0 when the ball reaches its maximum height. At that point, $t = 0.5$ s and $v_f = 0$:
$0 = v_i + at$
$0 = v_i + (-9.8)(0.5)$
$v_i = 4.9$
Now we can solve for $v_f$ using the equation we found for $v_i$ in terms of $v_f$:
$4.9 = -3v_f - 2$
$v_f = -2.3$
Finally, we can use the kinematic equation:
$v_f = v_i + at$
with $t = 0.5$ s and $a = -9.8$ m/s^2 to find the velocity when the ball passes the top of the flagpole:
$v_f = v_i + at$
$0 = v_i + (-9.8)(0.5)$
$v_i = 4.9$
Therefore, the speed of the ball as it passed the top of the flagpole was:
$|v_i| = |-4.9| = 4.9$
The answer is, therefore, (C) 17.6 m/s.

Question 17

An object is dropped from a bridge. A second object is thrown downwards 1.00 s later. They both reach the water 20.0 m below at the same instant. What was the initial speed of the second object?
A)4.91 m/s
B)14.6 m/s
C)9.90 m/s
D)19.6 m/s
E)21.3 m/s

Accepted Answer

The answer of An object is dropped from a bridge....

Question 18

A speeding car is traveling at a constant 30.0 m/s when it passes a stationary police car. If the police car delays its motion for 1.00 s before starting, what must the constant acceleration of the police car be to catch the speeding car after the police car travels a distance of 300 m? A)6.00 m/s² B)3.00 m/s² C)7.41 m/s² D)1.45 m/s² E)3.70 m/s²

Accepted Answer

To catch the speeding car, the police car must cover the same distance in the same time. Using the equations of motion, the police car's acceleration can be calculated as 7.41 m/s².

Question 19

The velocity of an object is given by the expression v(t) = 3.00 m/s + ( 2.00 m/s³)t². Determine the position of the object as a function of time if it is located at x = 1.00 m at time t = 0.00 s. A)( 2.00 m/s)t + 1.00 m B)(3.00 m/s)t + ( 0.667 m/s3)t³ C)( 2.00 m/s)t D)0.667 m E)1.00 m + (3.00 m/s)t + ( 0.667 m/s³)t³

Accepted Answer

The position of the object is given by the integral of the velocity with respect to time. Therefore, the position of the object is given by:$$x(t) = \int v(t) dt = \int (3.00 m/s + ( 2.00 m/s³)t²) dt = 3.00 m/s t + ( 0.667 m/s³)t³ + C$$where C is the constant of integration. To determine the value of C, we use the initial condition that the object is located at x = 1.00 m at time t = 0.00 s. Therefore, we have:$$x(0) = 1.00 m = 3.00 m/s (0) + ( 0.667 m/s³)(0)³ + C$$$$C = 1.00 m$$Therefore, the position of the object as a function of time is given by:$$x(t) = 3.00 m/s t + ( 0.667 m/s³)t³ + 1.00 m$$

Quiz 2: Describing Motion: Kinematics in One Dimension

To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of the pole after 0.5 s and then reaches the top of the pole again after a total elapsed time of 4.1 s. How high is the pole above the point where the ball was launched?

FIGURE 2-8 -Fig. 2-8 shows the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the total distance run by the player in the 10 s shown in the graph.

Two objects are dropped from a bridge, an interval of 1.00 s apart. What is their separation 1.00 s after the second object is released?

In a relay race, runner A is carrying the baton and has a speed of 3.4 m/s. When he is 25 m behind the starting line, runner B starts from rest and accelerates at 0.140 m/s². How fast is B traveling when A overtakes her?

An object is thrown upwards with a speed of 14 m/s. How long does it take to reach a height of 5.0 m above the projection point while descending?

FIGURE 2-8 -Fig. 2-8 shows the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the net displacement of the player for the 10 s shown on the graph.

A rock is dropped from a vertical cliff. The rock takes 7.00 s to reach the ground below the cliff. What is the height of the cliff?

The acceleration of an object as a function of time is given by a(t) = (3.00 m/s³) t. If the object has a velocity 1.00 m/s at time t = 1.00 s, what is the displacement of the object between time t = 2.00 s and time t = 4.00 s?

A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the car a deceleration of 3.50 m/s². If the car comes to a stop in a distance of 30.0 m, what was the car's original speed?

The acceleration of an object as a function of time is given by a(t) = (1.00 m/s²) t². If displacement of the object between time t = 1.00 s and time t = 2.00 s is 15.0 m, what is the velocity of the object at time t = 0.00 s?

The acceleration of an object as a function of time is given by a(t) = (3.00 m/s³) t. If the object is at rest at time t = 0.00 s, what is the velocity of the object at time t = 5.00 s?

If an object was freely falling, from what height would it need to be dropped to reach a speed of 70.0 m/s before reaching the ground?

A rock is dropped from a vertical cliff. The rock takes 3.00 s to reach the ground below the cliff. A second rock is thrown vertically from the cliff. It takes 2.00 s to reach the ground below the cliff from the time it is released. With what velocity was the second rock released?

A ball is dropped from somewhere above a window that is 2.00 m in height. As it falls, it is visible to a person looking through the window for 200 ms as it passes by the 2.00 m height of the window. From what height above the top of the window was the ball dropped?

In a relay race, runner A is carrying the baton and has a speed of 3.00 m/s. When he is 25.0 m behind the starting line, runner B starts from rest and accelerates at 0.100 m/s². How long afterwards will A catch up with B to pass the baton to B?

Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.50 s and reaches the level of the top of the pole after a total elapsed time of 4.10 s. What was the speed of the ball at as it passed the top of the flagpole?

An object is dropped from a bridge. A second object is thrown downwards 1.00 s later. They both reach the water 20.0 m below at the same instant. What was the initial speed of the second object?

A speeding car is traveling at a constant 30.0 m/s when it passes a stationary police car. If the police car delays its motion for 1.00 s before starting, what must the constant acceleration of the police car be to catch the speeding car after the police car travels a distance of 300 m?

The velocity of an object is given by the expression v(t) = 3.00 m/s + ( 2.00 m/s³) t². Determine the position of the object as a function of time if it is located at x = 1.00 m at time t = 0.00 s.