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 horizontal because it is released at the top of the Ferris wheel. The distance it travels horizontally is the circumference of the wheel, which is 2π(8.0m) = 50.3m. The time it takes for the ball to hit the ground is half of the period of the Ferris wheel, which is 5s. Using the equation d = vt + 0.5at^2 with vertical initial velocity of 0 and acceleration due to gravity of 9.81 m/s^2, we get:
d = 0 + 0.5(9.81 m/s^2)(5 s)^2 = 122.6 m
So the ball lands 122.6m away horizontally from the release point, and we can use the Pythagorean theorem to find the distance from the point on the ground directly under the release point:
√(8.0 m)^2 + (122.6 m)^2 ≈ 9.1 m

Question 2

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 centripetal acceleration is v^2/r = (4.0 m/s)^2 / 0.50 m = 32 m/s^2. At the top of the circle, the acceleration is directed towards the center of the circle, which is downwards.

Question 3

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 4

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...

Question 5

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 magnitude of the acceleration of an object in uniform circular motion is given by a = v^2/r, where v is the linear speed of the object and r is the radius of the circle. Since the length of the string remains the same, the radius of the circle also remains the same. When the speed is doubled, the magnitude of the acceleration becomes a' = (2v)^2/r = 4(v^2/r). Therefore, the magnitude of the acceleration becomes four times as great.

Question 6

A ferry boat is sailing at 12 km/h 30° 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°E of N
B)due N
C)30°W of N
D)45° E of N
E)none of these

Accepted Answer

The ferry's velocity relative to the shore is the vector sum of its velocity relative to the river and the river's velocity. The eastward river flow cancels out the westward component of the ferry's velocity, resulting in a net northward movement.

Question 7

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°
B)45°
C)60°
D)63°
E)90°

Accepted Answer

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

Question 8

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 9

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

Centripetal acceleration (a) is given by the formula a = v^2/r. Here, a = 25g = 25 * 9.8 m/s², and r = 1.0 m. Solving for v gives v = sqrt(a * r) = sqrt(25 * 9.8 * 1) = 15.65 m/s, which is approximately 16 m/s.

Question 10

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 11

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 12

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 13

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 formula for centripetal acceleration is 
a = v^2/r,
where a is the centripetal acceleration, v is the speed, and r is the radius of the curve.

Substituting the given values, we get:
a = (10 m/s)^2 / 20 m = 5 m/s^2

Therefore, the magnitude of the acceleration is 5 m/s^2. Choice (C) is the closest to this value. The other choices are either zero or too far off.

Question 14

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° E of N
B)30°E of N
C)45F° E of N
D)60° E of N
E)depends on the width of the river

Accepted Answer

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

Question 15

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 ° 
B)45 ° 
C)60 ° 
D)63 ° 
E)90 °

Accepted Answer

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

Question 16

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

First, we need to find the speed of the jogger. 
Speed = Distance/Time 
Speed = 25 m / 5 s 
Speed = 5 m/s

Next, we can find the circumference of the circle by multiplying the distance traveled by 4: 
Circumference = 25 m x 4 
Circumference = 100 m

Using the formula for centripetal acceleration: 
a = v^2 / r

We can rearrange the formula to solve for radius: 
r = v^2 / a

Plugging in the known values: 
r = (5 m/s)^2 / a 
r = 25 m^2 / a

We know that one fourth of the circumference is covered in 5 seconds, so we can set up the equation: 
(1/4) x 100 m = 25 m = distance traveled 
We also know that distance = circumference x fraction of circle traveled, so: 
25 m = (2πr / 4) 
25 m = πr / 2 
r = 50 m / π

Plugging this value of r into our earlier equation: 
50 m / π = 25 m^2 / a 
a = (25 m^2) / (50 m / π) 
a = 1.6 m/s^2

Therefore, the magnitude of her acceleration is 1.6 m/s^2.

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 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 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 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 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 ferry boat is sailing at 12 km/h 30° 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:

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.

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

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 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:

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

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: