Question 1

A Ferris wheel with a radius of 8.0m makes 1 revolution every 10 s. When he is at the top, essentially a diameter above the ground, he releases a ball. How far from the point on the ground directly under the release point does the ball land?
A) 0 m
B) 1.0 m
C) 8.0 m
D) 9.1 m
E) 16 m

Accepted Answer

The initial velocity of the ball is zero since it is released at the top of the Ferris wheel. The time for the ball to reach the ground can be found using the equation:
$\Delta y = v_{0}t + \frac{1}{2}at^{2}$
Since the ball falls under the influence of gravity, the acceleration is $-9.81 	ext{ m/s}^2$ (negative since it is downward).
The initial position of the ball is 8.0 m above the ground, so $\Delta y$ = -8.0 m.
The final position of the ball is at the point on the ground directly under the release point, so the final position is 0 m.
$v_{0}$ = 0 m/s.
Plugging in these values, we get:
$-8.0 	ext{ m} = 0 	ext{ m/s}(t) + \frac{1}{2}(-9.81 	ext{ m/s}^{2})(t)^{2}$
Solving for t, we get $t = \sqrt{\frac{16}{9.81}} \approx 1.28 	ext{ s}$.
The horizontal distance that the ball travels can be found using:
$\Delta x = v_{x}t$
where $v_{x}$ is the horizontal velocity of the ball. Since the ball is released at the top of the Ferris wheel, it has the same horizontal velocity as the Ferris wheel.
The distance that the Ferris wheel travels in one revolution is the circumference of the circle with radius 8.0 m:
$C = 2\pi r = 2\pi(8.0 	ext{ m}) \approx 50.27 	ext{ m}$
The time for one revolution is 10 s, so the angular velocity ($\omega$) of the Ferris wheel is:
$\omega = \frac{2\pi}{10 	ext{ s}} = 0.628 	ext{ rad/s}$
The horizontal velocity of the Ferris wheel is given by:
$v_{x} = r\omega = (8.0 	ext{ m})(0.628 	ext{ rad/s}) \approx 5.02 	ext{ m/s}$
Plugging in the values, we get:
$\Delta x = (5.02 	ext{ m/s})(1.28 	ext{ s}) \approx 6.43 	ext{ m}$
Therefore, the ball lands approximately 6.43 m from the point on the ground directly under the release point. The answer closest to this value is D) 9.1 m, which is the best choice.

Question 2

A girl wishes to swim across a river to a point directly opposite as shown. She can swim at 2 m/s in still water and the river is flowing at 1 m/s. At what angle $\theta$ with respect to the line joining the starting and finishing points should she swim? 
A) 30 $\degree$
B) 45 $\degree$
C) 60 $\degree$
D) 63 $\degree$
E) 90 $\degree$

Accepted Answer

The answer of A girl wishes to swim across a...

Question 3

A stone is tied to a string and whirled at constant speed in a horizontal circle. The speed is then doubled without changing the length of the string. Afterward the magnitude of the acceleration of the stone is:
A) the same
B) twice as great
C) four times as great
D) half as great
E) one-fourth as great

Accepted Answer

The acceleration of an object moving in a circle is given by a = v^2/r, where v is the speed and r is the radius of the circle. Since the length of the string remains constant, the radius of the circle also remains constant. When the speed is doubled, the new acceleration is a' = (2v)^2/r = 4(v^2/r). Therefore, the new acceleration is four times as great as the original acceleration.

Question 4

A car rounds a 20-m radius curve at a speed of 10 m/s. The magnitude of its acceleration is: A) 0 m/s² B) 0.20 m/s² C) 5.0 m/s² D) 40 m/s² E) 400 m/s²

Accepted Answer

The magnitude of the acceleration can be found using the formula a = v^2/r, where v is the velocity and r is the radius of curvature. Plugging in the values given, we get a = (10 m/s)^2 / 20 m = 5 m/s^2. Therefore, the answer is C, 5.0 m/s^2.

Question 5

A stone is tied to a 0.50-m string and whirled at a constant speed of 4.0 m/s in a vertical circle. The acceleration at the bottom of the circle is: A) 9.8 m/s², up B) 9.8 m/s², down C) 8.0 m/s², up D) 32 m/s², up E) 32 m/s², down

Accepted Answer

The answer of A stone is tied to a 0.50-m...

Question 6

An object is moving on a circular path of radius $\pi$ meters at a constant speed of 4.0 m/s. The time required for one revolution is: A) 2/ $\pi$² s B) $\pi$²/2 s C) $\pi$/2 s D) $\pi$²/4 s E) 2/ $\pi$ s

Accepted Answer

The time required for one revolution (period) is given by the formula T=2πr/v, where r is the radius of the circle and v is the constant speed. Plugging in the values, we get T=2π(F1F1F1S1πF1F1F10)/(4.0) = F1F1F1S1πF1F1F10/2 seconds. Therefore, the answer is B.

Question 7

A motor boat can travel at 10 km/h in still water. A river flows at 5 km/h west. A boater wishes to cross from the south bank to a point directly opposite on the north bank. At what angle must the boat be headed?
A) 27 $\degree$ E of N
B) 30 $\degree$ E of N
C) 45 $\degree$ E of N
D) 60 $\degree$ E of N
E) depends on the width of the river

Accepted Answer

The answer of A motor boat can travel at 10...

Question 8

For a biological sample in a 1.0-m radius centrifuge to have a centripetal acceleration of 25g its speed must be:
A) 11 m/s
B) 16 m/s
C) 50 m/s
D) 122 m/s
E) 245 m/s

Accepted Answer

The answer of For a biological sample in a 1.0-m...

Question 9

Two objects are traveling around different circular orbits with constant speed. They both have the same acceleration but object A is traveling twice as fast as object B. The orbit radius for object A is _______ the orbit radius for object B.
A) one-fourth
B) one-half
C) the same as
D) twice
E) four times

Accepted Answer

The centripetal acceleration formula is $a = \frac{v^2}{r}$, where $a$ is the acceleration, $v$ is the velocity, and $r$ is the radius of the circular path. Given that both objects have the same acceleration but object A is traveling twice as fast as object B ($v_A = 2v_B$), we can set their accelerations equal to each other: $\frac{(2v_B)^2}{r_A} = \frac{v_B^2}{r_B}$. Simplifying gives $\frac{4v_B^2}{r_A} = \frac{v_B^2}{r_B}$, which further simplifies to $r_A = 4r_B$. Thus, the orbit radius for object A is four times the orbit radius for object B.

Question 10

A stone is tied to the end of a string and is swung with constant speed around a horizontal circle with a radius of 1.5 m. If it makes two complete revolutions each second, its acceleration is: A) 0.24 m/s² B) 2.4 m/s² C) 24 m/s² D) 240 m/s² E) 2400 m/s²

Accepted Answer

The answer of A stone is tied to the end...

Question 11

A girl jogs around a horizontal circle with a constant speed. She travels one fourth of a revolution, a distance of 25 m along the circumference of the circle, in 5.0 s. The magnitude of her acceleration is: A) 0.31 m/s² B) 1.3 m/s² C) 1.6 m/s² D) 3.9 m/s² E) 6.3 m/s²

Accepted Answer

We can use the formula for linear acceleration, a = v^2/r, where v is the linear speed (circumference/time) and r is the radius of the circle. Since she travels one fourth of a revolution, her total distance traveled is one fourth of the circumference of the circle, so the circumference is 100 meters. Thus, her linear speed is 25/5 = 5 m/s. The radius can be found from the formula for the circumference of a circle, C = 2πr, so r = C/2π = 50/π meters. Plugging in these values gives:

a = v^2/r = (5 m/s)^2 / (50/π) m = 1.6 m/s^2

So the magnitude of her acceleration is 1.6 m/s^2, which is answer choice C.

Question 12

A boat is able to move through still water at 20 m/s. It makes a round trip to a town 3.0 km upstream. If the river flows at 5 m/s, the time required for this round trip is:
A) 120 s
B) 150 s
C) 200 s
D) 300 s
E) 320 s

Accepted Answer

The speed of the boat in still water is 20 m/s, and the speed of the river is 5 m/s. When going upstream, the effective speed of the boat is $20 - 5 = 15$ m/s, and when going downstream, it is $20 + 5 = 25$ m/s. The distance to the town is 3.0 km, which is 3000 m. The time to go upstream is $3000 m / 15 m/s = 200$ s, and the time to go downstream is $3000 m / 25 m/s = 120$ s. The total time for the round trip is $200 s + 120 s = 320$ s.

Question 13

A boy wishes to cross a river in the shortest possible time, without necessarily landing directly across from his starting point. He can row at 2 m/s in still water and the river is flowing at 1 m/s. At what angle $\theta$  should he point the bow (front) of his boat? 
A) 30 $\degree$
B) 45 $\degree$
C) 60 $\degree$
D) 63 $\degree$
E) 90 $\degree$

Accepted Answer

The answer of A boy wishes to cross a river...

Question 14

A stone is tied to a 0.50-m string and whirled at a constant speed of 4.0 m/s in a vertical circle. Its acceleration at the top of the circle is: A) 9.8 m/s², up B) 9.8 m/s², down C) 8.0 m/s², down D) 32 m/s², up E) 32 m/s², down

Accepted Answer

The answer of A stone is tied to a 0.50-m...

Question 15

A ferry boat is sailing at 12 km/h 30 $\degree$ W of N with respect to a river that is flowing at 6.0 km/h E. As observed from the shore, the ferry boat is sailing:
A) 30 $\degree$ E of N
B) due N
C) 30 $\degree$ W of N
D) 45 $\degree$E of N
E) none of these

Accepted Answer

The answer of A ferry boat is sailing at 12...

Question 16

A boat is traveling upstream at 14 mph with respect to a river that is flowing at 6 mph (with respect to the ground). A man runs directly across the boat, from one side to the other, at 6 mph (with respect to the boat). The speed of the man with respect to the ground is:
A) 10 mph
B) 14 mph
C) 18.5 mph
D) 21 mph
E) 26 mph

Accepted Answer

The answer of A boat is traveling upstream at 14...

Quiz 4: Motion in Two and Three Dimensions

A Ferris wheel with a radius of 8.0m makes 1 revolution every 10 s. When he is at the top, essentially a diameter above the ground, he releases a ball. How far from the point on the ground directly under the release point does the ball land?

A stone is tied to a string and whirled at constant speed in a horizontal circle. The speed is then doubled without changing the length of the string. Afterward the magnitude of the acceleration of the stone is:

A car rounds a 20-m radius curve at a speed of 10 m/s. The magnitude of its acceleration is:

A stone is tied to a 0.50-m string and whirled at a constant speed of 4.0 m/s in a vertical circle. The acceleration at the bottom of the circle is:

An object is moving on a circular path of radius $\pi$ meters at a constant speed of 4.0 m/s. The time required for one revolution is:

A motor boat can travel at 10 km/h in still water. A river flows at 5 km/h west. A boater wishes to cross from the south bank to a point directly opposite on the north bank. At what angle must the boat be headed?

For a biological sample in a 1.0-m radius centrifuge to have a centripetal acceleration of 25g its speed must be:

Two objects are traveling around different circular orbits with constant speed. They both have the same acceleration but object A is traveling twice as fast as object B. The orbit radius for object A is _______ the orbit radius for object B.

A stone is tied to the end of a string and is swung with constant speed around a horizontal circle with a radius of 1.5 m. If it makes two complete revolutions each second, its acceleration is:

A girl jogs around a horizontal circle with a constant speed. She travels one fourth of a revolution, a distance of 25 m along the circumference of the circle, in 5.0 s. The magnitude of her acceleration is:

A boat is able to move through still water at 20 m/s. It makes a round trip to a town 3.0 km upstream. If the river flows at 5 m/s, the time required for this round trip is:

A stone is tied to a 0.50-m string and whirled at a constant speed of 4.0 m/s in a vertical circle. Its acceleration at the top of the circle is:

A ferry boat is sailing at 12 km/h 30 $\degree$ W of N with respect to a river that is flowing at 6.0 km/h E. As observed from the shore, the ferry boat is sailing:

A boat is traveling upstream at 14 mph with respect to a river that is flowing at 6 mph (with respect to the ground) . A man runs directly across the boat, from one side to the other, at 6 mph (with respect to the boat) . The speed of the man with respect to the ground is: