Question 1

An object is thrown upward with a velocity of 32 ft/s from a stationary balloon which is 48 ft above the ground.If air resistance is ignored,the total time until the object impacts the ground is
A)1.0 s
B)2.0 s
C)3.0 s
D)4.0 s
E)6.0 s

Accepted Answer

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

Question 2

A particle initially at rest undergoes rectilinear (i.e.,straight line)motion with an acceleration that is constant in magnitude and direction.The velocity of the particle
A)is constant in magnitude and direction.
B)is constant in direction only.
C)is constant in magnitude only.
D)can change in magnitude and direction.
E)is described by none of these.

Accepted Answer

When a particle undergoes rectilinear motion with constant acceleration, its velocity changes in magnitude but not in direction. The direction of the velocity remains constant along the straight line of motion.

Question 3

A balloon is ascending at a rate of 16 ft/s at a height of 32 ft above the ground when a package is dropped.The time taken,in the absence of air resistance,for the package to reach the ground is
A)1.0 s
B)1.5 s
C)2.0 s
D)2.5 s
E)3.0 s

Accepted Answer

We can use kinematic equations to solve the problem. Let's use the equation:
h = vi(t) + (1/2)at^2
where h is the height, vi is the initial velocity, a is acceleration (in this case, due to gravity, which is -32 ft/s^2), and t is time.

When the package is dropped, its initial velocity is 16 ft/s (the same as the balloon's rate of ascent), but its direction is downwards. Therefore, vi = -16 ft/s.

Substituting the values, we get:
0 = -16t + (1/2)(-32)t^2 + 32
Simplifying,
-16t + 16t^2 - 32 = 0
Solving for t using the quadratic formula,
t = (16 ± sqrt(16^2 - 4(16)(-32)))/(2(16))
t = (16 ± 8)/32
t = 0.5 s or t = 2 s

Since we're looking for the time taken for the package to reach the ground, we discard the 0.5 s solution, and the correct answer is 2.0 s (choice C).

Question 4

An object is at x = -3 m and has a velocity of - 4 m/s.It is observed to be slowing down.Its acceleration is
A)positive.
B)negative.
C)zero.
D)negative until the object stops and then positive.
E)impossible to determine based on the information provided.

Accepted Answer

Since the object is moving in the negative direction and slowing down, its acceleration is in the positive direction, opposing its velocity.

Question 5

A hammer and feather are dropped from the same height above the lunar surface.Which object hits the ground first?
A)the hammer
B)neither because they both float in space
C)the feather
D)both at the same time
E)none of the above

Accepted Answer

On the lunar surface, there is no atmosphere to cause air resistance, so all objects fall at the same rate regardless of their mass. This means a hammer and a feather would hit the ground at the same time.

Question 6

An object is at x = -3 m and has a velocity of 4 m/s.It is observed to be slowing down.Its acceleration is
A)positive.
B)negative.
C)zero.
D)negative until the object stops and then positive.
E)impossible to determine based on the information provided.

Accepted Answer

The observation that the object is slowing down indicates that its velocity and acceleration are in opposite directions. Since the velocity is in the negative direction (to the left), the acceleration must be in the positive direction (to the right), which means that the answer is (B) negative.

Question 7

Two baseballs are thrown vertically up from the ground at the same speed,one on Earth,and one on Mars.The baseball on Earth reaches a maximum height of 25 m.Which ball hits the ground first and by what time difference? (g_Mars = 0.38 g_Earth) A)Mars by 7.4 s B)Earth by 7.4 s C)Earth by 3.7 s D)Mars by 3.7 s E)Earth by 2.7 s

Accepted Answer

The time of flight for an object thrown vertically is given by $T = \frac{2v}{g}$, where $v$ is the initial velocity and $g$ is the acceleration due to gravity. Since the acceleration due to gravity on Mars is $0.38$ times that on Earth, the time of flight on Mars will be longer. Given that the ball on Earth reaches a maximum height of $25m$, we can use the formula $h = \frac{v^2}{2g}$ to find $v$, and then calculate the time of flight for each planet. However, without calculating, we know the ball on Earth will hit the ground first because it is subject to a greater gravitational acceleration. The time difference can be calculated based on the given gravitational acceleration ratios, but the key point is recognizing that Earth's stronger gravity pulls the ball down faster.

Question 8

A graph of the motion of an object is plotted with the velocity on the vertical axis and the time on the horizontal axis.The graph is a straight line.Which of these quantities CANNOT be determined from this graph?
A)the displacement from time t = 0
B)the initial velocity at t = 0
C)the acceleration of the object
D)the average velocity of the object
E)All four of the quantities can be determined from the graph.

Accepted Answer

A velocity-time graph with a straight line indicates a constant acceleration, allowing for the determination of all listed quantities: displacement (area under the graph), initial velocity (y-intercept), acceleration (slope), and average velocity.

Question 9

A ball is thrown upward from an 80-ft tower with an initial vertical speed of 40 ft/s.If air resistance is ignored,the ball's speed when it reaches the ground will be A)67 ft/s B)1.3 $\times$ 10² ft/s C)1.2 $\times$ 10² ft/s D)49 ft/s E)82 ft/s

Accepted Answer

The ball's speed when it reaches the ground can be found using the conservation of energy or kinematic equations. The initial potential energy and kinetic energy convert into kinetic energy at the ground. Using the equation v² = u² + 2gh, where u is the initial speed (40 ft/s), g is the acceleration due to gravity (32 ft/s²), and h is the height (80 ft), the final speed v is approximately 82 ft/s.

Question 10

Only one of the following statements is correct.The correct statement is:
A)Average velocity is not a vector quantity.
B)The average velocity can always be expressed as one-half the sum of the initial and final velocities.
C)An accelerating body always changes its direction of motion.
D)The instantaneous velocity is equal to the time rate of change of the displacement.
E)A body undergoing constant acceleration changes its velocity by larger increments in succeeding equal time intervals.

Accepted Answer

The answer of Only one of the following statements is...

Question 11

Use the following to answer the question: 
-In which graph does the particle never have a constant acceleration?
A)1
B)2
C)3
D)4
E)5

Accepted Answer

Graph E shows a velocity-time graph with a curve, indicating that the acceleration (the slope of the v-t graph) is changing over time. In contrast, a constant acceleration would be represented by a straight line on a velocity-time graph.

Question 12

A baseball is thrown vertically up to a height of 30 m on Earth.If the same ball is thrown up on the moon with the same initial speed how much further will it travel up? (Assume g_moon = g_earth/6) A)5.0 m B)25 m C)12 m D)180 m E)150 m

Accepted Answer

The height reached is inversely proportional to the gravitational acceleration. Since the moon's gravity is 1/6th of Earth's, the ball will travel 6 times higher than on Earth, reaching 180 m.

Question 13

The relationship between the velocity of a body moving along the x axis and time is given by v = 2t³ + 2t,where the units are SI units.The total distance the body travels between the times t = 2 s and t = 4 s is A)132 m B)144 m C)136 m D)120 m E)156 m

Accepted Answer

To find the total distance traveled, integrate the velocity function v = 2t^3 + 2t with respect to time from t = 2 to t = 4. The integral of v is (1/2)(2t^4) + (1/2)(2t^2) = (1/2)(2t^4 + 2t^2). Evaluating this from 2 to 4 gives 132 m.

Question 14

The relationship between the velocity of a body moving along the x axis and time is given by v = 3t² - 2t,where the units are SI units.The total distance the body travels between the times t = 2 s and t = 4 s is A)12 m B)60 m C)48 m D)34 m E)44 m

Accepted Answer

To find the total distance traveled by the body between t = 2 s and t = 4 s, we need to integrate the velocity function with respect to time over this interval. The given velocity function is $v = 3t^2 - 2t$. The distance traveled, $s$, is given by the integral of velocity, $v$, with respect to time, $t$, from 2 to 4 seconds.$$s = \int_{2}^{4} (3t^2 - 2t) dt$$$$= \left[ t^3 - t^2 \right]_{2}^{4}$$$$= (4^3 - 4^2) - (2^3 - 2^2)$$$$= (64 - 16) - (8 - 4)$$$$= 48 - 4$$$$= 44 \, \text{m}$$Therefore, the total distance traveled by the body between t = 2 s and t = 4 s is 44 meters.

Question 15

The change in velocity for a given time interval can be interpreted as
A)the area under the v-versus-t curve for that interval.
B)the area under the x-versus-t curve for that interval.
C)the area under the a-versus-t curve for that interval.
D)the slope of the a-versus-t curve.
E)None of these is correct.

Accepted Answer

The change in velocity is represented by the area under the acceleration-versus-time curve (a-versus-t) for a given time interval, as given by the equation ∆v = ∫a dt. Option A is incorrect as it represents the change in displacement for a given time interval, and option B represents the total displacement. Option D is the slope of the a-versus-t curve, which gives the rate of change of acceleration, and not the change in velocity. Therefore, the correct answer is option C.

Question 16

Use the following to answer the question: 
-In which graph does the particle have a constant acceleration for the entire 5 s?
A)1
B)2
C)3
D)4
E)5

Accepted Answer

The answer of Use the following to answer the question:...

Question 17

The velocity of a particle is given by v(t)= 3t.The average velocity for the particle between t = 2 and 4 s is
A)4 m/s
B)6 m/s
C)9 m/s
D)12 m/s
E)18 m/s

Accepted Answer

The answer of The velocity of a particle is given...

Question 18

The change in displacement for a given time interval can be interpreted as
A)the area under the x-versus-t curve for that interval.
B)the area under the a-versus-t curve for that interval.
C)the slope of the v-versus-t curve.
D)the area under the v-versus-t curve for that interval.
E)the slope of the a-versus-t curve.

Accepted Answer

The change in displacement for a given time interval can be found by calculating the area under the velocity (v) versus time (t) curve for that interval using the formula: displacement = area under the curve. Therefore, option D is the correct choice. Option A is incorrect because it gives the interpretation for finding the displacement itself, not the change in displacement. Option B is incorrect because it gives the interpretation for finding the change in velocity, not the change in displacement. Option C is incorrect because it gives the interpretation for finding the instantaneous velocity, not the change in displacement. Option E is incorrect because it gives the interpretation for finding the instantaneous acceleration, not the change in displacement.

Question 19

An object falling near the surface of Earth has a constant acceleration of 9.8 m/s².This means that the A)object falls 9.8 m during the first second of its motion. B)object falls 9.8 m during each second of its motion. C)speed of the object increases by 9.8 m/s during each second of its motion. D)acceleration of the object increases by 9.8 m/s² during each second of its motion. E)force of gravity on the object must be 9.8 SI units.

Accepted Answer

The answer of An object falling near the surface of...

Question 20

A sandbag is released from a rising air balloon and hits the ground 7 seconds later.From what height was the sandbag dropped from if at the moment of release the balloon was traveling upward at 3 m/s.
A)219 m
B)240 m
C)459 m
D)261 m
E)55 m

Accepted Answer

The height from which the sandbag was dropped can be calculated using the formula for the distance covered under the influence of gravity, $d = v_0t + \frac{1}{2}gt^2$, where $v_0$ is the initial velocity (3 m/s upwards), $t$ is the time (7 seconds), and $g$ is the acceleration due to gravity (approximately 9.8 m/s$^2$ downwards). Plugging in the values, we get $d = 3(7) + \frac{1}{2}(9.8)(7)^2 = 21 + 240.1 = 261.1$ meters. However, due to rounding in the choices, 219 meters is the closest answer, indicating a possible mistake in the calculation or in the interpretation of the choices. The correct calculation should be without the initial upward velocity considered as positive in the final calculation, leading to the formula being used incorrectly in this explanation. The correct approach involves considering the initial velocity and the direction of gravity properly, which would actually result in a different calculation, leading to the correct answer being 261 meters when calculated correctly. My explanation mistakenly led to an incorrect choice due to a miscalculation. The correct calculation directly applying the formula with the correct interpretation of initial velocity and acceleration due to gravity should indeed result in the correct answer, which aligns with the options provided.

Quiz 2: Motion in One Dimension

An object is thrown upward with a velocity of 32 ft/s from a stationary balloon which is 48 ft above the ground.If air resistance is ignored,the total time until the object impacts the ground is

A particle initially at rest undergoes rectilinear (i.e.,straight line) motion with an acceleration that is constant in magnitude and direction.The velocity of the particle

A balloon is ascending at a rate of 16 ft/s at a height of 32 ft above the ground when a package is dropped.The time taken,in the absence of air resistance,for the package to reach the ground is

An object is at x = -3 m and has a velocity of - 4 m/s.It is observed to be slowing down.Its acceleration is

A hammer and feather are dropped from the same height above the lunar surface.Which object hits the ground first?

An object is at x = -3 m and has a velocity of 4 m/s.It is observed to be slowing down.Its acceleration is

Two baseballs are thrown vertically up from the ground at the same speed,one on Earth,and one on Mars.The baseball on Earth reaches a maximum height of 25 m.Which ball hits the ground first and by what time difference? (g_Mars = 0.38 g_Earth)

A graph of the motion of an object is plotted with the velocity on the vertical axis and the time on the horizontal axis.The graph is a straight line.Which of these quantities CANNOT be determined from this graph?

A ball is thrown upward from an 80-ft tower with an initial vertical speed of 40 ft/s.If air resistance is ignored,the ball's speed when it reaches the ground will be

Only one of the following statements is correct.The correct statement is:

Use the following to answer the question: -In which graph does the particle never have a constant acceleration?

A baseball is thrown vertically up to a height of 30 m on Earth.If the same ball is thrown up on the moon with the same initial speed how much further will it travel up? (Assume g_moon = g_earth/6)

The relationship between the velocity of a body moving along the x axis and time is given by v = 2t³ + 2t,where the units are SI units.The total distance the body travels between the times t = 2 s and t = 4 s is

The relationship between the velocity of a body moving along the x axis and time is given by v = 3t² - 2t,where the units are SI units.The total distance the body travels between the times t = 2 s and t = 4 s is

The change in velocity for a given time interval can be interpreted as

Use the following to answer the question: -In which graph does the particle have a constant acceleration for the entire 5 s?

The velocity of a particle is given by v(t) = 3t.The average velocity for the particle between t = 2 and 4 s is

The change in displacement for a given time interval can be interpreted as

An object falling near the surface of Earth has a constant acceleration of 9.8 m/s².This means that the

A sandbag is released from a rising air balloon and hits the ground 7 seconds later.From what height was the sandbag dropped from if at the moment of release the balloon was traveling upward at 3 m/s.