Question 1

A package is dropped from a helicopter moving upward at 15 m/s.If it takes 16.0 s before the package strikes the ground,how high above the ground was the package when it was released? (Disregard air resistance. )
A)1000 m
B)1500 m
C)810 m
D)1200 m

Accepted Answer

The package has an initial upward velocity of 15 m/s and an acceleration of -9.8 m/s^2 due to gravity. The time it takes to reach the ground is 16.0 s. We can use the following equation to find the initial height of the package:h = vi*t + 1/2*a*t^2h = 15 m/s * 16.0 s + 1/2*(-9.8 m/s^2)*(16.0 s)^2h = 240 m + 1228.8 mh = 1468.8 mTherefore, the package was released from a height of 1468.8 m above the ground.

Question 2

At the same moment,one rock is dropped and one is thrown downward with an initial velocity of 10 m/s from the top of a 300 m building.How much earlier does the thrown rock strike the ground? (Disregard air resistance. )
A)0.95 s
B)0.66 s
C)0.85 s
D)They land at exactly the same time.

Accepted Answer

The time taken for the dropped rock is calculated using the equation $ h = \frac{1}{2}gt^2 $, which gives $ t \approx 7.75 $ seconds. For the thrown rock, use $ h = v_i t + \frac{1}{2}gt^2 $, which gives $ t \approx 6.8 $ seconds. The difference is approximately 0.95 seconds.

Question 3

It is physically impossible for an object to have a negative acceleration and yet be speeding up.

Accepted Answer

Acceleration is a vector quantity, which means it has both magnitude and direction. If an object is speeding up, it means that its velocity is increasing, which implies that the object has acceleration in the same direction as its velocity vector. If the acceleration is negative, it means that it is acting in the opposite direction to the velocity vector, which cannot cause an increase in speed. Therefore, an object cannot have a negative acceleration and be speeding up at the same time.

Question 4

The equation sf = si + vis Δt + as (Δt)² is valid for all types of motion because it is a fundamental equation of physics.

Accepted Answer

The answer of The equation sf = si + vis...

Question 5

A ball is projected upward at time t = 0.0 s,from a point on a roof 90 m above the ground.The ball rises,then falls and strikes the ground.The initial velocity of the ball is 80.5 m/s.Consider all quantities as positive in the upward direction.The velocity of the ball when it is 89 m above the ground is closest to:
A)- 81 m/s
B)- 64 m/s
C)- 48 m/s
D)- 32 m/s
E)- 97 m/s

Accepted Answer

Explanation:The velocity of the ball when it is 89 m above the ground is closest to -81 m/s. This is because the ball is moving upward at a speed of 80.5 m/s when it is projected, and it will continue to move upward until it reaches its maximum height. At its maximum height, the ball will have a velocity of 0 m/s, and it will then begin to fall back to the ground. As it falls, its velocity will increase until it reaches the ground, at which point it will have a velocity of -80.5 m/s.

Question 6

A ball is projected upward at time t = 0.0 s,from a point on a roof 60 m above the ground.The ball rises,then falls and strikes the ground.The initial velocity of the ball is 28.4 m/s.Consider all quantities as positive in the upward direction.At time t = 4.3 s,the acceleration of the ball is closest to: A)zero B)+5 m/s² C)+10 m/s² D)-5 m/s² E)-10 m/s²

Accepted Answer

At time t=4.3s, the ball has already reached its maximum height and is on its way down. Therefore, its acceleration is -g, which is approximately -10 m/s^2 (assuming no air resistance). Since all quantities are positive in the upward direction, the acceleration is considered negative, so the closest choice is E, -10 m/s^2.

Question 7

Is there any situation in which the average velocity of an object can be greater than the object's average speed? Support your answer.

Accepted Answer

The answer of Is there any situation in which the...

Question 8

If the "velocity versus time" graph of an object is a horizontal line,that object cannot be accelerating.

Accepted Answer

Since acceleration is the rate of change of velocity, if the velocity is constant (as indicated by a horizontal line on the graph), the acceleration is zero. Therefore, the object cannot be accelerating.

Question 9

A toy rocket is launched vertically from ground level (y = 0 m),at time t = 0.0 s.The rocket engine provides constant upward acceleration during the burn phase.At the instant of engine burnout,the rocket has risen to 81 m and acquired a velocity of 40 m/s.The rocket continues to rise in unpowered flight,reaches maximum height,and falls back to the ground.The upward acceleration of the rocket during the burn phase is closest to: A)9.9 m/s² B)9.6 m/s² C)9.3 m/s² D)9.0 m/s² E)8.7 m/s²

Accepted Answer

Using the kinematic equation $ v^2 = u^2 + 2as $, where $ v = 40 \, \text{m/s} $, $ u = 0 \, \text{m/s} $, and $ s = 81 \, \text{m} $, we solve for $ a $ to find the acceleration during the burn phase.

Question 10

A toy rocket is launched vertically from ground level (y = 0 m),at time t = 0.0 s.The rocket engine provides constant upward acceleration during the burn phase.At the instant of engine burnout,the rocket has risen to 49 m and acquired a velocity of 60 m/s.The rocket continues to rise in unpowered flight,reaches maximum height,and falls back to the ground.The maximum height reached by the rocket is closest to:
A)233 m
B)221 m
C)209 m
D)244 m
E)256 m

Accepted Answer

Using the kinematic equation $v^2 = u^2 + 2as$, where $v = 0$ at maximum height, $u = 60 \, \text{m/s}$, and $a = -9.8 \, \text{m/s}^2$, we find the additional height after burnout is 184 m. Adding the initial 49 m gives a total height of 233 m.

Question 11

It is not physically possible for the "position versus time" graph of a moving animal to be either perfectly vertical or perfectly horizontal.

Accepted Answer

A perfectly vertical line would imply infinite speed, and a perfectly horizontal line would imply no movement, which contradicts the definition of a moving animal.

Question 12

What quantity is measured by your speedometer (be as specific as possible)?

Accepted Answer

The answer of What quantity is measured by your speedometer...

Question 13

The equation vfs² = vis²+ 2a_s Δx applies to motion for which the "velocity versus time" graph is a straight line.

Accepted Answer

The equation given is for uniformly accelerated motion, where the velocity changes at a constant rate. The "velocity versus time" graph for uniformly accelerated motion is not a straight line, but a curve. Therefore, the statement in the question is false.

Question 14

A test rocket is fired straight up from rest with a net acceleration of 20 m/s².After 4 seconds the motor turns off,but the rocket continues to coast upward.What maximum elevation does the rocket reach? A)487 m B)327 m C)320 m D)408 m E)160 m

Accepted Answer

Explanation:The rocket reaches its maximum elevation when its velocity is zero. We can use the equation of motion:v = u + atwhere v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.At the maximum elevation, v = 0, so we have:0 = u + at-u = att = -u/aWe know that the initial velocity is 0, so we have:t = 0This means that the rocket reaches its maximum elevation at the instant the motor turns off.We can now use the equation of motion:s = ut + 1/2 at^2where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.At the maximum elevation, s is the maximum elevation, u is 0, and t is 4 seconds, so we have:s = 1/2 * a * t^2s = 1/2 * 20 m/s^2 * (4 s)^2s = 160 mTherefore, the maximum elevation is 160 m.

Question 15

A ball is thrown straight upward with a velocity of 39 m/s.How much time passes before the ball strikes the ground? (Disregard air resistance. )
A)8.0 s
B)4.0 s
C)2.4 s
D)1.2 s

Accepted Answer

The time to reach the highest point is given by t = v/g = 39 m/s / 9.8 m/s² = 4 s. The total time for the round trip is twice this, so 8 s.

Question 16

If an object stops moving at a point,then its acceleration must be zero at that point.

Accepted Answer

An object can stop moving at a point due to a change in direction or velocity, but it can still have acceleration if there is a force acting on it, such as gravity or friction.

Question 17

Negative acceleration is called deceleration because an object is slowing down.

Accepted Answer

Negative acceleration refers to acceleration in the opposite direction of the velocity, which may not always mean slowing down. It depends on the direction of the velocity vector.

Question 18

You go on a long trip and try to determine your average velocity by using the miles on your car's tripometer and the time the trip required.Why would the answer using this information most likely not be correct?

Accepted Answer

The answer of You go on a long trip and...

Question 19

The acceleration is always the slope of the "velocity versus time" graph and the velocity is always the slope of the "position versus time" graph.

Accepted Answer

This statement is true according to the fundamental concept of calculus, where the derivative of a function represents its instantaneous rate of change. Therefore, the slope of the velocity versus time graph gives us the acceleration, and the slope of the position versus time graph gives us the velocity.

Question 20

A sports car can go from rest to 32 m/s in 3.88 s.The same car can come to a full stop from that speed in 3.96 s.What is the ratio of starting to stopping accelerations?
A)-1.0
B)1.0
C)-0.98
D)0.98

Accepted Answer

The answer of A sports car can go from rest...

Quiz 2: Motion in One Dimension

A package is dropped from a helicopter moving upward at 15 m/s.If it takes 16.0 s before the package strikes the ground,how high above the ground was the package when it was released? (Disregard air resistance. )

At the same moment,one rock is dropped and one is thrown downward with an initial velocity of 10 m/s from the top of a 300 m building.How much earlier does the thrown rock strike the ground? (Disregard air resistance. )

It is physically impossible for an object to have a negative acceleration and yet be speeding up.

The equation sf = si + vis Δt + as (Δt)2 is valid for all types of motion because it is a fundamental equation of physics.

Is there any situation in which the average velocity of an object can be greater than the object's average speed? Support your answer.

If the "velocity versus time" graph of an object is a horizontal line,that object cannot be accelerating.

It is not physically possible for the "position versus time" graph of a moving animal to be either perfectly vertical or perfectly horizontal.

What quantity is measured by your speedometer (be as specific as possible)?

The equation vfs2 = vis2+ 2as Δx applies to motion for which the "velocity versus time" graph is a straight line.

A test rocket is fired straight up from rest with a net acceleration of 20 m/s2.After 4 seconds the motor turns off,but the rocket continues to coast upward.What maximum elevation does the rocket reach?

A ball is thrown straight upward with a velocity of 39 m/s.How much time passes before the ball strikes the ground? (Disregard air resistance. )