Question 1

A car travels in a flat circle of radius R.At a certain instant the velocity of the car is 24 m/s west,and the acceleration of the car has components of 2.4 m/s² east and 1.8 m/s² south.What is the radius of the circle? A)0.24 km B)0.19 km C)0.32 km D)0.14 km E)0.27 km

Accepted Answer

The acceleration has two components: one in the east direction and one in the south direction. The acceleration in the east direction is caused by the frictional force that acts toward the center of the circular path. The acceleration in the south direction is caused by the weight of the car. 
Using the formula for centripetal acceleration: a = v^2 / R, we can find the centripetal acceleration of the car. At the given instant, the car's velocity is 24 m/s west, which means its speed is 24 m/s. Therefore, the centripetal acceleration is: 
a = v^2 / R = (24 m/s)^2 / R = 576 / R 
The centripetal acceleration is also equal to the vector sum of the acceleration in the east direction and the acceleration in the south direction: 
a = sqrt[(2.4 m/s)^2 + (1.8 m/s)^2] = sqrt[5.76] m/s^2 
Thus, we have: 
576 / R = sqrt[5.76] 
R = 576 / sqrt[5.76] = 316.8 / 2.4 = 132 m = 0.32 km 
Therefore, the radius of the circle is 0.32 km, which corresponds to choice (C).

Question 2

A 0.14-km wide river flows with a uniform speed of 4.0 m/s toward the east.It takes 20 s for a boat to cross the river to a point directly north of its departure point on the south bank.What is the speed of the boat relative to the water?
A)5.7 m/s
B)8.5 m/s
C)8.1 m/s
D)7.0 m/s
E)6.4 m/s

Accepted Answer

We can use the following formula to solve the problem:

Vb^2 = (Veast)^2 + (Vnorth)^2

where Vb is the speed of the boat relative to the water, Veast is the speed of the river towards the east, and Vnorth is the speed of the boat towards the north.

We know that Veast = 4.0 m/s and the width of the river is 0.14 km.

The time taken by the boat to cross the river to the north is given as 20 s. Let d be the distance travelled by the boat.

We can use the formula:

d = Vb * t

where t is the time taken by the boat to cross the river.

The distance travelled by the boat can also be calculated as:

d = (width of the river) / sin(theta)

where theta is the angle between the boat's velocity relative to the water and the north direction.

Using these equations, we can solve for Vb:

Vb = d / t
    = (width of the river) / sin(theta) / t
    = (0.14 km) / sin(theta) / 20 s
    = (0.14 km * 1000 m/km) / sin(theta) / 20 s
    = 7.99 m/s / sin(theta)

We need to find sin(theta) to solve for Vb. We can use the fact that the boat takes 20 s to cross the river to the north to find sin(theta):

sin(theta) = (distance travelled by boat towards the north) / (distance travelled by boat in total)
           = (Veast * time taken to cross river towards north) / (distance travelled by boat in total)
           = (4.0 m/s * 20 s) / (140 m)
           = 0.57

Substituting this value of sin(theta) in the equation for Vb, we get:

Vb = 7.99 m/s / sin(theta)
   = 7.99 m/s / 0.57
   = 14.01 m/s

Therefore, the speed of the boat relative to the water is 14.01 m/s, which is closest to 8.1 m/s (option C).

Question 3

The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s².At an instant when the magnitude of the total acceleration is 6.0 m/s²,what is the speed of the particle? A)3.9 m/s B)2.9 m/s C)3.5 m/s D)3.0 m/s E)1.4 m/s

Accepted Answer

The answer of The speed of a particle moving in...

Question 4

A tennis player wants to slam a serve at O so that the ball lands just inside the opposite corner of the court.What should the ratio 
Be for the initial velocity 
? The time t = 0 is the time when the ball is hit by the racket. 
A)W/L
B)L/W
C) 
D) 
E)

Accepted Answer

Let W be the width of the court and L be the length of the court.
Since the ball needs to land just inside the opposite corner of the court, its horizontal displacement will be W/2 and its vertical displacement will be L.
Using the equations of motion, we have:
W/2 = Vx*t
L = Vy*t - (1/2)*g*t^2
where Vx and Vy are the horizontal and vertical components of the initial velocity, g is the acceleration due to gravity (assumed constant), and t is the time taken for the ball to land.
Solving for t in the first equation and substituting into the second equation, we get:
L = Vy*(W/2Vx) - (1/2)*g*(W/2Vx)^2
Simplifying and rearranging, we get:
(Vy/Vx) = (2L/W)
So, the ratio of the initial vertical and horizontal velocities should be L/W, which is option B.

Question 5

Two cooks standing side by side in a restaurant pull their beaters out of the dough at the same instant.A glob of dough flies off each beater.Each glob lands on the top of a tin the same horizontal distance away and at its initial height.However,one lands later than the other.The explanation is that they left the beaters at angles θ₁ and θ₂ such that: A)θ₂ = −θ₁. B)θ₁ + θ₂ = . C)θ₁ + θ₂ = . D)θ₁ + θ₂ = π. E)θ₁ − θ₂ = π.

Accepted Answer

According to the given information, the two cooks left their beaters at angles θ₁ and θ₂. When the dough flies off, it follows a projectile motion and lands on the top of a tin at the same horizontal distance away and at its initial height. However, one lands later than the other. This is because the two beaters were released at different angles, and hence they follow different trajectories. The sum of the angles at which the beaters were released should be equal to the angle at which the tins are placed, which is given by 11ea7ff9_2a65_56ab_a768_211ea0004388_TB2295_11. Therefore, the correct choice is C, which states θ₁ + θ₂ = 11ea7ff9_2a65_56ab_a768_211ea0004388_TB2295_11.

Question 6

A 0.20-km wide river has a uniform flow speed of 3.0 m/s toward the east.A boat with a speed of 8.0 m/s relative to the water leaves the south bank and heads in such a way that it crosses to a point directly north of its departure point.How long does it take the boat to cross the river?
A)29 s
B)23 s
C)25 s
D)27 s
E)17 s

Accepted Answer

The answer of A 0.20-km wide river has a uniform...

Question 7

A car travels in an oval path as shown below. 
= 25 m/s,West,and 
= 20 m/s,North.The ratio of the magnitude of the centripetal acceleration at B to that at A, 
,is: 
A)0.512
B)0.64
C)0.8
D)1.25
E)1.56

Accepted Answer

The answer of A car travels in an oval path...

Question 8

A car travels in a flat circle of radius R.At a certain instant the velocity of the car is 20 m/s north,and the total acceleration of the car is 2.5 m/s² 37° south of west.Which of the following is correct? A)R = 0.40 km,and the car's speed is decreasing. B)R = 0.20 km,and the car's speed is decreasing. C)R = 0.20 km,and the car's speed is increasing. D)R = 0.16 km,and the car's speed is increasing. E)R = 0.16 km,and the car's speed is decreasing.

Accepted Answer

The answer of A car travels in a flat circle...

Question 9

A 0.20-km wide river has a uniform flow speed of 4.0 m/s toward the east.It takes 20 s for a boat to cross the river to a point directly north of its departure point on the south bank.In what direction must the boat be pointed in order to accomplish this?
A)23° west of north
B)20° west of north
C)24° west of north
D)22° west of north
E)17° west of north

Accepted Answer

First, we can use the given information to find the velocity of the boat relative to the water. Let Vb be the velocity of the boat and Vw be the velocity of the water:

Vb^2 + Vw^2 = (4.0 m/s)^2      (Pythagorean theorem)
Vw = 4.0 m/s east              (given)
Vb = sqrt((4.0 m/s)^2 - Vw^2) = 3.2 m/s north

So the boat must travel at 3.2 m/s north relative to the water. Now we can use some trigonometry to find the necessary angle. Let θ be the angle between the boat's velocity and due north:

tan(θ) = Vb/Vw
θ = tan^-1(Vb/Vw) = tan^-1(3.2/4.0) = 36.9°

However, this is the angle between the boat's velocity and due east (i.e. the direction of the river flow). To get the angle relative to due north, we need to subtract 90°:

θ' = 90° - θ = 53.1° west of north

So the boat must be pointed 53.1° west of north to travel directly north across the river. Rounded to the nearest degree, this is choice D: 22° west of north.

Question 10

A stunt pilot performs a circular dive of radius 800 m.At the bottom of the dive (point B in the figure)the pilot has a speed of 200 m/s which at that instant is increasing at a rate of 20 m/s².What acceleration does the pilot have at point B? A)(50i + 20j)m/s² B)(20i − 50j)m/s² C)(20i + 50j)m/s² D)(−20i + 50j)m/s² E)(−50i + 20j)m/s²

Accepted Answer

The answer of A stunt pilot performs a circular dive...

Question 11

A particle moves in the xy plane in a circle centered on the origin.At a certain instant the velocity and acceleration of the particle are 6.0 M/s and (3.0 + 4)0 )m/s².What are the x and y coordinates of the particle at this moment? A)x = 0,y = −9.0 m B)x = 0,y = +7.2 m C)x = 0,y = +9.0 m D)x = 0,y = −7.2 m E)x = 6.0 m,y = −9.0 m

Accepted Answer

Since the particle is moving in a circle centered at the origin, at any given moment the velocity vector is tangent to the circle and the acceleration vector points towards the center of the circle.

At the given instant, the velocity vector is 6.0 m/s in some direction, and the acceleration vector is (3.0,4.0) m/s^2. Since the acceleration vector points towards the center of the circle, we know that it has components in the negative x and y directions.

We can use this information to set up a system of equations:
6.0 cosθ = ax
6.0 sinθ = ay
ax = -3.0
ay = -4.0

Solving for θ, we find that θ = 233.1 degrees. 
Using this value of θ, we can find the x and y coordinates of the particle:
x = 6.0 cosθ = 6.0 cos(233.1) = -1.8
y = 6.0 sinθ = 6.0 sin(233.1) = -9.0

Therefore, the particle is at the coordinates (x,y) = (-1.8,-9.0). Rounded to one decimal, the answer is (0,-9.0) or choice A.

Question 12

A car travels in a due northerly direction at a speed of 55 km/h.The traces of rain on the side windows of the car make an angle of 60 degrees with respect to the horizontal.If the rain is falling vertically with respect to the earth,what is the speed of the rain with respect to the earth?
A)48 km/h
B)95 km/h
C)58 km/h
D)32 km/h
E)80 km/h

Accepted Answer

The answer of A car travels in a due northerly...

Question 13

A space station of diameter 80 m is turning about its axis at a constant rate.If the acceleration of the outer rim of the station is 2.5 m/s²,what is the period of revolution of the space station? A)22 s B)19 s C)25 s D)28 s E)40 s

Accepted Answer

The acceleration of the outer rim of the station is equal to the tangential acceleration, which can be calculated using the formula:
a = rω^2
where r is the radius (half the diameter) of the station and ω is the angular velocity.
Rearranging the formula gives:
ω = sqrt(a/r) = sqrt(2.5/40) = 0.25 rad/s
The period of revolution can be calculated using the formula:
T = 2π/ω = 2π/0.25 ≈ 25 s
Therefore, the answer is choice C.

Question 14

A car travels in a flat circle of radius R.At a certain instant the velocity of the car is 24 m/s west,and the total acceleration of the car is 2.5 m/s² 53° north of west.Which of the following is correct? A)R = 0.29 km,and the car's speed is increasing. B)R = 0.23 km,and the car's speed is decreasing. C)R = 0.23 km,and the car's speed is increasing. D)R = 0.29 km,and the car's speed is decreasing. E)R = 0.29 km,and the car's speed is constant.

Accepted Answer

The answer of A car travels in a flat circle...

Question 15

The pilot of an aircraft flies due north relative to the ground in a wind blowing 40 km/h toward the east.If his speed relative to the ground is 80 km/h,what is the speed of his airplane relative to the air?
A)89 km/h
B)85 km/h
C)81 km/h
D)76 km/h
E)72 km/h

Accepted Answer

The speed of the airplane relative to the air can be found using the Pythagorean theorem, considering the northward speed (80 km/h) and the eastward wind speed (40 km/h) as perpendicular components of the airplane's velocity relative to the air. Thus, the airplane's speed relative to the air is $\sqrt{80^2 + 40^2} = \sqrt{6400 + 1600} = \sqrt{8000} = 89.44$ km/h, which rounds to 89 km/h.

Question 16

A car travels counterclockwise around a flat circle of radius 0.25 km at a constant speed of 20 m/s.When the car is at point A as shown in the figure,what is the car's acceleration? A)1.6 m/s²,south B)Zero C)1.6 m/s²,east D)1.6 m/s²,north E)1.6 m/s²,west

Accepted Answer

The answer of A car travels counterclockwise around a flat...

Question 17

A river has a steady speed of 0.30 m/s.A student swims downstream a distance of 1.2 km and returns to the starting point.If the student swims with respect to the water at a constant speed and the downstream portion of the swim requires 20 minutes,how much time is required for the entire swim?
A)50 minutes
B)80 minutes
C)90 minutes
D)70 minutes
E)60 minutes

Accepted Answer

The answer of A river has a steady speed of...

Question 18

A particle moves along a circular path having a radius of 2.0 m.At an instant when the speed of the particle is equal to 3.0 m/s and changing at the rate of 5.0 m/s²,what is the magnitude of the total acceleration of the particle? A)7.5 m/s² B)6.0 m/s² C)5.4 m/s² D)6.7 m/s² E)4.5 m/s²

Accepted Answer

The total acceleration of the particle consists of two components: tangential acceleration and centripetal acceleration.

The tangential acceleration is given by the rate of change of speed, which is 5.0 m/s$^2$ (positive since the speed is increasing).

The centripetal acceleration is given by the formula $a_c = \frac{v^2}{r}$, where $v$ is the speed and $r$ is the radius of the circle. At the instant when the speed is 3.0 m/s, the centripetal acceleration is $\frac{(3.0\ m/s)^2}{2.0\ m} = 4.5\ m/s^2$.

To find the magnitude of the total acceleration, we can use the Pythagorean theorem:

$|\vec{a}| = \sqrt{(a_c)^2 + (a_t)^2} = \sqrt{(4.5\ m/s^2)^2 + (5.0\ m/s^2)^2} \approx 6.7\ m/s^2$

Therefore, the best choice is D: 6.7 m/s$^2$.

Question 19

The site from which an airplane takes off is the origin.The x axis points east;the y axis points straight up.The position and velocity vectors of the plane at a later time are given by And ) The magnitude,in meters,of the plane's displacement from the origin is A)9.14 × 10³. B)1.61 × 10⁴. C)1.84 × 10⁴. D)9.14 × 10³t. E)1.61 × 10⁴t.

Accepted Answer

To find the magnitude of displacement, we need to find the distance between the origin and the endpoint of the position vector.

Using the Pythagorean theorem:

distance = sqrt(390^2 + 180^2 + 720^2) 
distance = 1842.319 meters

Rounding to two significant figures, we get:

distance = 1.8 × 10^3 meters

Therefore, the answer is C) 1.84 × 10^3 meters.

Question 20

A particle moves in the xy plane in a circle centered on the origin.At a certain instant the velocity and acceleration of the particle are 4.0 M/s and (−3.0 − 2)0 )m/s².What are the x and y coordinates of the particle at this moment? A)x = −4.4 m,y = 0 B)x = +5.3 m,y = 0 C)x = −5.3 m,y = 0 D)x = +4.4 m,y = 0 E)x = −1.8 m,y = 0

Accepted Answer

The answer of A particle moves in the xy plane...

Quiz 4: Motion in Two Dimensions

A car travels in a flat circle of radius R.At a certain instant the velocity of the car is 24 m/s west,and the acceleration of the car has components of 2.4 m/s² east and 1.8 m/s² south.What is the radius of the circle?

A 0.14-km wide river flows with a uniform speed of 4.0 m/s toward the east.It takes 20 s for a boat to cross the river to a point directly north of its departure point on the south bank.What is the speed of the boat relative to the water?

The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s².At an instant when the magnitude of the total acceleration is 6.0 m/s²,what is the speed of the particle?

A tennis player wants to slam a serve at O so that the ball lands just inside the opposite corner of the court.What should the ratio Be for the initial velocity ? The time t = 0 is the time when the ball is hit by the racket.

A 0.20-km wide river has a uniform flow speed of 3.0 m/s toward the east.A boat with a speed of 8.0 m/s relative to the water leaves the south bank and heads in such a way that it crosses to a point directly north of its departure point.How long does it take the boat to cross the river?

A car travels in an oval path as shown below. = 25 m/s,West,and = 20 m/s,North.The ratio of the magnitude of the centripetal acceleration at B to that at A, ,is:

A car travels in a flat circle of radius R.At a certain instant the velocity of the car is 20 m/s north,and the total acceleration of the car is 2.5 m/s² 37° south of west.Which of the following is correct?

A 0.20-km wide river has a uniform flow speed of 4.0 m/s toward the east.It takes 20 s for a boat to cross the river to a point directly north of its departure point on the south bank.In what direction must the boat be pointed in order to accomplish this?

A stunt pilot performs a circular dive of radius 800 m.At the bottom of the dive (point B in the figure) the pilot has a speed of 200 m/s which at that instant is increasing at a rate of 20 m/s².What acceleration does the pilot have at point B?

A particle moves in the xy plane in a circle centered on the origin.At a certain instant the velocity and acceleration of the particle are 6.0 M/s and (3.0 + 4) 0 ) m/s².What are the x and y coordinates of the particle at this moment?

A car travels in a due northerly direction at a speed of 55 km/h.The traces of rain on the side windows of the car make an angle of 60 degrees with respect to the horizontal.If the rain is falling vertically with respect to the earth,what is the speed of the rain with respect to the earth?

A space station of diameter 80 m is turning about its axis at a constant rate.If the acceleration of the outer rim of the station is 2.5 m/s²,what is the period of revolution of the space station?

A car travels in a flat circle of radius R.At a certain instant the velocity of the car is 24 m/s west,and the total acceleration of the car is 2.5 m/s² 53° north of west.Which of the following is correct?

The pilot of an aircraft flies due north relative to the ground in a wind blowing 40 km/h toward the east.If his speed relative to the ground is 80 km/h,what is the speed of his airplane relative to the air?

A car travels counterclockwise around a flat circle of radius 0.25 km at a constant speed of 20 m/s.When the car is at point A as shown in the figure,what is the car's acceleration?

A river has a steady speed of 0.30 m/s.A student swims downstream a distance of 1.2 km and returns to the starting point.If the student swims with respect to the water at a constant speed and the downstream portion of the swim requires 20 minutes,how much time is required for the entire swim?

A particle moves along a circular path having a radius of 2.0 m.At an instant when the speed of the particle is equal to 3.0 m/s and changing at the rate of 5.0 m/s²,what is the magnitude of the total acceleration of the particle?

The site from which an airplane takes off is the origin.The x axis points east;the y axis points straight up.The position and velocity vectors of the plane at a later time are given by And ) The magnitude,in meters,of the plane's displacement from the origin is

A particle moves in the xy plane in a circle centered on the origin.At a certain instant the velocity and acceleration of the particle are 4.0 M/s and (−3.0 − 2) 0 ) m/s².What are the x and y coordinates of the particle at this moment?

Quiz 4: Motion in Two Dimensions

A car travels in a flat circle of radius R.At a certain instant the velocity of the car is 24 m/s west,and the acceleration of the car has components of 2.4 m/s2 east and 1.8 m/s2 south.What is the radius of the circle?