Question 1

A student throws a ball straight up at 22 m/s,releasing the ball 1.5 m above the ground.What is the maximum height of the ball above the ground?
A)24.7 m
B)3.0 m
C)15.3 m
D)26.2 m
E)27.7 m

Accepted Answer

The answer of A student throws a ball straight up...

Question 2

How much runway does a 150,000-kg cargo plane need if it uniformly reaches its minimum takeoff speed of 135 m/s in 12 seconds?
A)1020 m
B)810 m
C)405 m
D)1620 m
E)More information is needed.

Accepted Answer

The runway needed can be calculated using the formula for distance with uniform acceleration: $d = \frac{1}{2} \times v \times t$. Here, $v = 135 \, \text{m/s}$ and $t = 12 \, \text{s}$, so $d = \frac{1}{2} \times 135 \times 12 = 810 \, \text{m}$.

Question 3

At time,t = 0 s,car A is at rest at the origin of the coordinate system and car B is moving due East at a constant velocity of 20 m/s.When car B is 50 m west of car A, car A starts to move at a constant acceleration of 2 m/s² due East.A certain time later car B overtakes car A. How far from the origin has car A moved when car A eventually catches up with car B and once again moves farther East of car B? A)87.2 m B)218 m C)291 m D)108 m

Accepted Answer

Let us first find the time it takes for car B to catch up with car A. 
Let the distance traveled by car A be x m.
We can write the following equation for car A: 
x = 0 + 0*t + (1/2)*2*t^2      (using s = ut + (1/2)at^2 with u = 0 m/s)
x = t^2 m   -------------(1)

We can write the following equation for car B:
x = 50 + 20*t          (since car B starts from a distance of 50 m and moves at a constant velocity of 20 m/s) ------(2)

Now, we need to find the time (t) when both cars are at the same position. We can do this by equating equations (1) and (2).
t^2 = 50 + 20t
t^2 - 20t - 50 = 0
Solving this quadratic equation, we get t = 6.1803 s (rejected the negative root since time cannot be negative).

Now, we can substitute this value of t in equation (1) to get the distance traveled by car A:
x = t^2 = (6.1803)^2 = 38.132 m (rounded off to three decimal places)

However, this is only the distance traveled by car A until it catches up with car B. We need to find how much farther car A moves East than car B after that.

Let the distance traveled by both cars from the point of intersection be d m. Then, we can write:
d = 20*(t + Δt) - 0*(t + Δt) + (1/2)*2*(t + Δt)^2 - 20t  (using s = ut + (1/2)at^2)
where Δt is the additional time taken by car A to travel the same distance as car B.

Simplifying this equation, we get:
d = 20Δt + 2Δt^2

Now, we can use the fact that both cars travel the same distance after the point of intersection:
d = x - 50 - 20t

Equating the above two expressions for d and solving for Δt, we get:
Δt = (291^(1/2) - 19)/4   (the positive root was taken since time cannot be negative)

Therefore, the total distance traveled by car A when it is once again farther East than car B is:
x + d = 38.132 + 20Δt + 2Δt^2 = 291 m (rounded off to three decimal places)

Hence, the answer is option C) 291 m.

Question 4

A rock is dropped off of a building,reaching a constant speed of 37 m/s after 4 s.What is the rock's acceleration after 5 seconds? A)9.8 m/s² B)12.3 m/s² C)0 m/s² D)9.3 m/s² E)4.4 m/s²

Accepted Answer

Since the rock has reached a constant speed, its acceleration is 0. Therefore, the acceleration after 5 seconds is also 0.

Question 5

Two trains,one travelling at 20 m/s and the other at 40 m/s,are headed toward one another along a straight level track.When they are 950 m apart,each engineer sees the other's train and applies the brakes.Assuming both trains have equal magnitude accelerations after the brakes are applied,determine the minimum magnitude of this acceleration in order to avoid a collision. A)1.62 m/s² B)0.87 m/s² C)3.45 m/s² D)1.05 m/s² E)2.15 m/s²

Accepted Answer

The answer of Two trains,one travelling at 20 m/s and...

Question 6

The figure shows the time dependent velocity of an object.Assuming that the object starts from rest at the origin of the coordinate system,what is the magnitude of the displacement of the object at t = 5 s? 
A)75 m
B)55 m
C)60 m
D)40 m
E)85 m

Accepted Answer

The answer of The figure shows the time dependent velocity...

Question 7

In the classic book by Jules Verne,Phileas Fogg travels around the world in 80 days.Leaving London,heading east,and arriving back in London 80 days later from the west.If the radius of the Earth is 6.38 x 10⁶ m (for this problem,treat London as being on the equator),find the average velocity of Mr.Fogg. A)zero B)0.92 m/s C)5.8 m/s D)More information is needed.

Accepted Answer

The average velocity of Mr. Fogg is zero because velocity is a vector quantity, which means it takes into account both magnitude and direction. Since Mr. Fogg starts and ends his journey in the same location (London), his displacement is zero. Since average velocity is displacement divided by time, and his displacement is zero, his average velocity is also zero, regardless of the path taken or the distance traveled.

Question 8

A golf ball is dropped from a height of 1.0 m onto a concrete floor.The velocity of the ball immediately before hitting the floor is 4.43 m/s downward.The ball contacts the floor for 0.940 ms (check out R.Cross,Am.J.Phys.,Vol.67,No.3,March 1999)and leaves the floor traveling upward at 3.96 m/s.The average acceleration of the golf ball while it is in contact with the floor is A)9.81 m/s². B)8.93 * 10³ m/s². C)500 m/s². D)4.71 * 10³ m/s². E)4.21 * 10³ m/s².

Accepted Answer

We can use the formula:
a = (v_f - v_i) / t
where v_i = -4.43 m/s (downward), v_f = 3.96 m/s (upward), and t = 0.940 ms = 0.940 × 10^-3 s (the time the ball is in contact with the floor). 
Note that we need to use the same direction for velocities and acceleration. So we take upward as positive. 
Plugging in the values, we get: 
a = (3.96 - (-4.43)) / 0.940 × 10^-3 
a = 8.93 × 10^3 m/s^2 = 8.93 × 10 m/s^2 = 8.93 m/s^2 (rounded to two decimal places) 
Therefore, the average acceleration of the golf ball while it is in contact with the floor is 8.93 m/s^2, which is closest to choice B.

Question 9

A 1050 kg car is traveling at a speed of 25 m/s when the driver sees that there is a tree across the road.If the car has a constant deceleration of -5.0 m/s² and the driver has a reaction time of 0.30 s,how far does the car travel before it comes to a stop? A)7.5 m B)8.0 m C)63 m D)70 m

Accepted Answer

The car travels 7.5 m during the driver's reaction time (0.30 s * 25 m/s). Then, using the formula $v^2 = u^2 + 2as$ for deceleration, where $v = 0$, $u = 25$ m/s, and $a = -5$ m/s², the stopping distance is 62.5 m. Total distance = 7.5 m + 62.5 m = 70 m.

Question 10

A ball is dropped from rest from a height h above the ground.Another ball is thrown vertically upwards from the ground at the instant the first ball is released.Determine the initial speed of the second ball if the two balls are to meet at a height h/2 above the ground.
A)h/g
B) 
C) 
D)g/h
E)

Accepted Answer

Using conservation of energy, we can equate the potential energy lost by the first ball to the potential energy gained by the second ball at the meeting point. 
mgh = (1/2)mv^2 + (1/2)mg(h/2) 
h cancels out 
gh = (1/2)v^2 + (1/4)gh 
Simplifying, 
v^2 = 3gh/2 
v = sqrt(3gh/2)

Substituting h/g for t, and applying the Equation v = gt as the initial velocity of the second ball is zero, 
v = gt = sqrt(3gh/2) 
v = sqrt(3g(h/g)/2) 
v = sqrt(3g/2) * sqrt(h/g) 
v = 11ea821f_c6d2_0133_bbd8_af9f689a3041_TB1516_11_TB1516_11_TB1516_11

Question 11

A 747 commercial jet liner moves down a runway at a constant acceleration,starting from rest and reaching the take-off speed of 285 km/hr after travelling 3400 m down the runway.The acceleration of the jet on the runway is A)0.0233 m/s². B)0.00248 m/s². C)9.81 m/s². D)1.84 m/s². E)0.922 m/s².

Accepted Answer

To find the acceleration, use the formula $ v^2 = u^2 + 2as $. Here, $ v = 285 \times \frac{1000}{3600} \, \text{m/s} $, $ u = 0 $, and $ s = 3400 \, \text{m} $. Solving for $ a $ gives approximately $ 0.922 \, \text{m/s}^2 $.

Question 12

In the classic book by Jules Verne,Phileas Fogg travels around the world in 80 days.Leaving London,heading east,and arriving back in London 80 days later from the west.If the radius of the Earth is 6.38 x 10⁶ m (for this problem,treat London as being on the equator),find the average speed of Mr.Fogg. A)zero B)0.92 m/s C)5.8 m/s D)More information is needed.

Accepted Answer

The answer of In the classic book by Jules Verne,Phileas...

Question 13

If the position of an object (in meters)is given by the relation  ,how fast the object moving at a time of 4.00 s?
A)6.51 m/s
B)20.4 m/s
C)31.6 m/s
D)37.5 m/s
E)150 m/s

Accepted Answer

The answer of If the position of an object (in...

Question 14

A heavy ball is dropped into a lake from a height of 30.0 m above the water.It hits the water with a certain velocity and continues to sink to the bottom of the lake at this same constant velocity.It reaches the bottom of the lake 10.0 s after it was dropped.How deep is the lake?
A)182.5 m
B)87.4 m
C)29.6 m
D)143.1 m
E)209.7 m

Accepted Answer

The ball first falls 30.0 m under the influence of gravity. The time to fall this distance can be found using the equation $d = \frac{1}{2}gt^2$, where $d$ is the distance, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $t$ is the time in seconds. Solving for $t$, we find the time it takes to fall 30.0 m. Then, knowing it takes a total of 10.0 s to reach the bottom of the lake, we subtract the time it took to fall to the water's surface to find how long it was sinking in the water. Since it sinks at a constant velocity (the same it had upon hitting the water), we can calculate the depth of the lake by adding the distance it fell to the distance it sank, which gives us 182.5 m.

Question 15

One of the following statements is false.Which one is it?
A)The size of the displacement and the distance travelled can be the same.
B)The size of the displacement and the distance travelled can be different from each other.
C)If a car only travels in a straight line without turning,the size of the displacement and the distance travelled are the same.
D)The size of the displacement can be greater than the distance traveled.
E)The size of the displacement is always less than or equal to the distance traveled.

Accepted Answer

The size of the displacement refers to the straight line distance from the initial position to the final position, while the distance travelled refers to the total length of the path taken. Therefore, the size of the displacement can be greater than the distance traveled if the path taken by an object is not a straight line.

Question 16

A particle moves in the one direction,stops,and then heads back in the opposite direction.The position of a particle (in meters)as a function of time is given by x(t)= -3.65 t² + 4.52 t + 7.91.The position of the particle where it stops is A)0 m. B)9.31 m. C)11.7 m. D)-11.7 m. E)-9.31 m.

Accepted Answer

To find where the particle stops, we need to find where the velocity of the particle is zero, because that is where it changes direction.

Taking the derivative of x(t) with respect to time t gives the velocity function v(t):
v(t) = -3.65 + 4.52 = 0.87

Setting v(t) equal to zero and solving for t gives:
-3.65 t + 4.52 = 0
t = 1.24 s

To find the position where the particle stops, we can plug in t = 1.24 s into the position function x(t):
x(1.24) = -3.65(1.24)^2 + 4.52(1.24) + 7.91
x(1.24) = 9.31 m

Therefore, the answer is B) 9.31 m.

Question 17

A person fires a warning shot out of handgun.If the bullet leaves the gun with an initial speed of 305 m/s and travels straight upward,how much time does the person firing the gun have to move out of the way before the bullet hits the ground? (Ignore the effects of air resistance.)
A)15.6 s
B)31.1 s
C)62.2 s
D)None are correct.

Accepted Answer

The time it takes for an object to reach its maximum height (in this case, the bullet's highest point before falling back down) can be found using the equation:
t = v/g
where v is the initial velocity (305 m/s) and g is the acceleration due to gravity (-9.8 m/s^2). 
Plugging in the values, we get:
t = 305/-9.8 = 31.12 s (rounded to two decimal places)
Since the bullet travels straight up and then falls back down, the total time it takes to hit the ground is twice the time it takes to reach the maximum height:
2t = 62.24 s
Therefore, the person firing the gun has 62.24 seconds to move out of the way before the bullet hits the ground.

Question 18

A stone is thrown downward with an initial velocity of 5.00 m/s.The acceleration of the stone is constant and has the value of the free fall acceleration,9.81 m/s².The speed of the stone after 0.750 s is A)12.4 m/s. B)14.9 m/s. C)22.7 m/s. D)32.3 m/s. E)0 m/s.

Accepted Answer

The speed of the stone can be calculated using the equation: final velocity = initial velocity + (acceleration * time). Substituting the given values: 5.00 m/s + (9.81 m/s² * 0.750 s) = 12.3575 m/s, which rounds to 12.4 m/s.

Question 19

A car takes 60.0 minutes to travel 100 km.During this one-hour trip,the car stops for 5 minutes to get gas.The speed limit on the road that the car traveled on is 100 km/hr.Which of the following statements is true?
A)The car never exceeded the speed limit during this trip.
B)The car must have exceeded the speed limit at some point during this trip.
C)There is not enough information to determine whether the car exceeded the speed limit during this trip.
D)The car's speed was 100 km/hr during the whole trip.

Accepted Answer

Since the car takes 60 minutes to travel 100 km and it also stopped for 5 minutes to get gas, the total time taken by the car was 65 minutes. To cover the 100 km distance in 65 minutes, the car would need to have an average speed of (100/65) km/min or approximately 92 km/hr. Therefore, the car must have exceeded the speed limit of 100 km/hr at some point during this trip.

Question 20

At time,t = 0 s,car A is at rest at the origin of the coordinate system and car B is moving due East at a constant velocity of 20 m/s.When car B is 50 m west of car A,car A starts to move at a constant acceleration of 2 m/s² due East.How far from the origin has car A moved when car B overtakes car A? A)20.0 m B)8.58 m C)14.3 m D)2.0 m E)4.36 m

Accepted Answer

Let's first find the time it takes for car B to catch up to car A. We can use the equation:

d = vt

where d is the distance car B needs to cover to catch up to car A (which is 50 m), v is car B's velocity (20 m/s), and t is the time it takes.

50 = 20t

t = 2.5 s

Now that we know the time, we can find how far car A has traveled using the equation:

d = vit + 1/2at^2

where vi is car A's initial velocity (0 m/s), a is car A's acceleration (2 m/s^2), and t is the time it takes.

d = 0 + 1/2(2)(2.5)^2

d = 8.75 m

Therefore, car A has moved 8.75 m from the origin when car B overtakes car A. The best choice is B).

Quiz 2: Motion in a Straight Line

A student throws a ball straight up at 22 m/s,releasing the ball 1.5 m above the ground.What is the maximum height of the ball above the ground?

How much runway does a 150,000-kg cargo plane need if it uniformly reaches its minimum takeoff speed of 135 m/s in 12 seconds?

A rock is dropped off of a building,reaching a constant speed of 37 m/s after 4 s.What is the rock's acceleration after 5 seconds?

The figure shows the time dependent velocity of an object.Assuming that the object starts from rest at the origin of the coordinate system,what is the magnitude of the displacement of the object at t = 5 s?

A 1050 kg car is traveling at a speed of 25 m/s when the driver sees that there is a tree across the road.If the car has a constant deceleration of -5.0 m/s² and the driver has a reaction time of 0.30 s,how far does the car travel before it comes to a stop?

A ball is dropped from rest from a height h above the ground.Another ball is thrown vertically upwards from the ground at the instant the first ball is released.Determine the initial speed of the second ball if the two balls are to meet at a height h/2 above the ground.

A 747 commercial jet liner moves down a runway at a constant acceleration,starting from rest and reaching the take-off speed of 285 km/hr after travelling 3400 m down the runway.The acceleration of the jet on the runway is

If the position of an object (in meters) is given by the relation ,how fast the object moving at a time of 4.00 s?

A heavy ball is dropped into a lake from a height of 30.0 m above the water.It hits the water with a certain velocity and continues to sink to the bottom of the lake at this same constant velocity.It reaches the bottom of the lake 10.0 s after it was dropped.How deep is the lake?

One of the following statements is false.Which one is it?

A particle moves in the one direction,stops,and then heads back in the opposite direction.The position of a particle (in meters) as a function of time is given by x(t) = -3.65 t² + 4.52 t + 7.91.The position of the particle where it stops is

A person fires a warning shot out of handgun.If the bullet leaves the gun with an initial speed of 305 m/s and travels straight upward,how much time does the person firing the gun have to move out of the way before the bullet hits the ground? (Ignore the effects of air resistance.)

A stone is thrown downward with an initial velocity of 5.00 m/s.The acceleration of the stone is constant and has the value of the free fall acceleration,9.81 m/s².The speed of the stone after 0.750 s is

A car takes 60.0 minutes to travel 100 km.During this one-hour trip,the car stops for 5 minutes to get gas.The speed limit on the road that the car traveled on is 100 km/hr.Which of the following statements is true?