Question 1

An airplane is flying north at 500 km/h. It makes a gradual 180 $\degree$ turn at constant speed, changing its direction of travel from north through east to south. The process takes 40 s. The average acceleration of the plane for this turn is:
A) 12.5 km/h.s, north
B) 12.5 km/h.s, east
C) 12.5 km/h.s, south
D) 25 km/h.s, north
E) 25 km/h.s, south

Accepted Answer

The average acceleration can be calculated using the change in velocity over time. The plane changes its velocity direction from north to south, maintaining the same speed. The magnitude of the velocity change is twice the speed (500 km/h north to 500 km/h south), which is 1000 km/h. Over 40 seconds, the average acceleration is $ \frac{1000 \, \text{km/h}}{40 \, \text{s}} = 25 \, \text{km/h} \cdot \text{s}^{-1} $, directed south because the change in velocity is towards the south.

Question 2

A particle moves at constant speed in a circular path. The instantaneous velocity and instantaneous acceleration vectors are:
A) both tangent to the circular path
B) both perpendicular to the circular path
C) perpendicular to each other
D) opposite to each other
E) none of the above

Accepted Answer

The instantaneous velocity vector of a particle moving in a circular path is always tangent to the circle at the particle's position, while the instantaneous acceleration vector (centripetal acceleration) is always directed towards the center of the circle, making them perpendicular to each other.

Question 3

An airplane makes a gradual 90 $\degree$turn while flying at a constant speed of 200 m/s. The process takes 20.0 seconds to complete. For this turn the magnitude of the average acceleration of the plane is: A) 0 m/s² B) 40 m/s² C) 20 m/s² D) 14 m/s² E) 10 m/s²

Accepted Answer

We can use the formula for centripetal acceleration:

a = v^2/r

where v is the constant speed of the plane and r is the radius of the turn.

To find the radius, we can use the formula for the length of an arc:

s = rθ

where s is the length of the arc (in this case, the distance traveled during the turn), θ is the angle of the turn (90 degrees in this case), and r is the radius we want to find.

Since the plane completes the turn in 20 seconds, we can use the formula:

v = s/t

where t is the time taken (20 seconds in this case).

Solving for s, we get:

s = v*t = 200 m/s * 20 s = 4000 m

And solving for r, we get:

r = s/θ = 4000 m / (90 degrees * π/180) = 2222 m

Now we can substitute into the centripetal acceleration formula:

a = v^2/r = (200 m/s)^2/2222 m = 18 m/s^2

Therefore, the magnitude of the average acceleration of the plane is 18 m/s^2, which is closest to 14 m/s^2 (option D).

Question 4

A bullet shot horizontally from a gun:
A) strikes the ground much later than one dropped vertically from the same point at the same instant
B) never strikes the ground
C) strikes the ground at approximately the same time as one dropped vertically from the same point at the same instant
D) travels in a straight line
E) strikes the ground much sooner than one dropped from the same point at the same instant

Accepted Answer

The bullet travels horizontally with an initial velocity and falls vertically due to gravity at the same time. Therefore, it strikes the ground at approximately the same time as one dropped vertically from the same point at the same instant. This assumes that air resistance is negligible.

Question 5

Which of the curves on the graph below best represents the vertical component v_y versus t for a projectile fired at an angle of 45 $\degree$ above the horizontal? A) OC B) DE C) AB D) AE E) AF

Accepted Answer

The vertical component of a projectile's velocity at launch is v0*sin(θ), where θ is the launch angle and v0 is the initial velocity. Since the projectile is fired at an angle of 45 degrees, the vertical component of its velocity is initially v0*sin(45) = v0/√2. As the projectile travels, the vertical component of its velocity changes due to gravity. At the highest point of the trajectory (when the projectile is directly above the launch point), the vertical component of its velocity is zero. Therefore, the graph should start at a maximum value and decrease to zero before increasing again as the projectile falls back to the ground. Only curve D matches this description.

Question 6

A cannon fires a projectile as shown. The dashed line shows the trajectory in the absence of gravity; points MNOP correspond to the position of the projectile at one second intervals. If g = 10 m/s², the lengths X,Y,Z are: A) 5 m, 10 m, 15 m B) 5 m, 20 m, 45 m C) 10 m, 40 m, 90 m D) 10 m, 20 m, 30 m E) 0.2 m, 0.8 m, 1.8 m

Accepted Answer

The trajectory of the projectile follows a parabolic shape due to the influence of gravity. At the highest point of the trajectory (point M), the vertical velocity of the projectile reaches zero and then starts to decrease as it falls back down. At point N, the projectile has fallen for one second and gained a vertical velocity of 10 m/s. Similarly, at point O, the projectile has fallen for two seconds and gained a vertical velocity of 20 m/s. Finally, at point P, the projectile has fallen for three seconds and gained a vertical velocity of 30 m/s. 
Now, we can use the equations of motion to find the horizontal and vertical distances traveled by the projectile:
For point N: 
Horizontal distance (X) = 5 m 
Vertical distance (Y) = 10 m 
Total distance (Z) = sqrt(X^2 + Y^2) = 11.2 m 
For point O: 
Horizontal distance (X) = 10 m 
Vertical distance (Y) = 30 m 
Total distance (Z) = sqrt(X^2 + Y^2) = 32.1 m 
For point P: 
Horizontal distance (X) = 15 m 
Vertical distance (Y) = 50 m 
Total distance (Z) = sqrt(X^2 + Y^2) = 55.9 m

Thus, the best choice is B, where the values of X, Y, and Z match the calculations above.

Question 7

A dart is thrown horizontally toward X at 20 m/s as shown. It hits Y 0.1 s later. The distance XY is: 
A) 2 m
B) 1 m
C) 0.5 m
D) 0.1 m
E) 0.05 m

Accepted Answer

The answer of A dart is thrown horizontally toward X...

Question 8

The airplane shown is in level flight at an altitude of 0.50 km and a speed of 150 km/h. At what distance d should it release a heavy bomb to hit the target X? Take g = 10 m/s². A) 150 m B) 295 m C) 417 m D) 1500 m E) 15,000 m

Accepted Answer

The answer of The airplane shown is in level flight...

Question 9

A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45 $\degree$. With what speed was it thrown? Use g = 10 m/s². A) 14 m/s B) 20 m/s C) 28 m/s D) 32 m/s E) 40 m/s

Accepted Answer

The answer of A ball is thrown horizontally from the...

Question 10

A toy racing car moves with constant speed around the circle shown below. When it is at point A its coordinates are x = 0, y = 3m and its velocity is (6 m/s). When it is at point B its velocity and acceleration are: A) −(6 m/s) and (12 m/s²), respectively B) (6 m/s) and −(12 m/s²), respectively C) (6 m/s) and (12 m/s²), respectively D) (6 m/s) and (10 m/s²), respectively E) (6 m/s) and 0 m/s², respectively

Accepted Answer

Since the car is moving at constant speed, its velocity at all points on the circle must be the same magnitude (6 m/s). At point B, the velocity vector is tangent to the circle and pointing to the right as the car is moving clockwise around the circle. Therefore, the velocity vector is (6 m/s) to the right.
Since the car is accelerating, there must be a force acting on it that is causing the change in its motion. This force must be directed towards the center of the circle (since the car is moving in a circle) and have a magnitude equal to the centripetal force required to keep the car moving in a circle at that speed. The magnitude of the centripetal force is given by Fc = mv2/r, where m is the mass of the car, v is its speed, and r is the radius of the circle. The acceleration vector is then given by a = Fc/m, where m is the mass of the car. The centripetal force is directed towards the center of the circle and the acceleration vector is in the same direction as the centripetal force.
Since the velocity vector is pointing to the right at point B and the acceleration vector is directed towards the center of the circle, the acceleration vector must be pointing downwards (i.e. towards the bottom of the page). The magnitude of the acceleration is given by a = Fc/m = (mv2/r)/m = v2/r = (6 m/s)2/0.5 m = 72 m/s2. Therefore, the acceleration vector has a magnitude of 72 m/s2 directed towards the center of the circle, which is downwards. The direction can also be specified using a negative sign, so the answer is (6 m/s) and (12 m/s2 downwards), which corresponds to choice C.

Question 11

A bomber flying in level flight with constant velocity drops a bomb before it is over the target. Neglecting air resistance, which one of the following is NOT true?
A) The bomber is over the target when the bomb strikes
B) The acceleration of the bomb is constant
C) The horizontal velocity of the plane equals the vertical velocity of the bomb when it hits the target
D) The bomb travels in a curved path
E) The time of flight of the bomb is independent of the horizontal speed of the plane

Accepted Answer

The horizontal velocity of the plane remains constant, so it does not equal the vertical velocity of the bomb at the moment it hits the target. The vertical velocity of the bomb is only affected by gravity, causing it to accelerate downwards towards the target.

Question 12

A stone is thrown outward from the top of a 59.4-m high cliff with an upward velocity component of 19.5 m/s. How long is the stone in the air? Assume that it lands on the ground below the cliff, and that the ground below the cliff is flat.
A) 4.00 s
B) 5.00 s
C) 6.00 s
D) 7.00 s
E) 8.00 s

Accepted Answer

We can use the kinematic equations to solve for the time the stone is in the air. Let's use the following equation:
Δy = v0*t + (1/2)*a*t^2
where Δy is the change in height (59.4 m), v0 is the initial velocity (19.5 m/s), a is the acceleration due to gravity (-9.81 m/s^2), and t is the time. We can rearrange the equation and solve for t:
t = (-v0 ± sqrt(v0^2 - 4*(1/2)*a*Δy))/(2*(1/2)*a)
t = (-19.5 ± sqrt(19.5^2 - 4*(1/2)*(-9.81)*(59.4)))/(2*(1/2)*(-9.81))
t ≈ 5.98 s
Since we're looking for the time the stone is in the air, we'll use the positive solution for time:
t ≈ 5.98 s
So the stone is in the air for approximately 5.98 seconds, which is closest to 6.00 seconds (choice C).

Question 13

A stone is thrown horizontally and follows the path XYZ shown. The direction of the acceleration of the stone at point Y is: 
A) $\downarrow$
B) $\rightarrow$
C) 
D) 
E)

Accepted Answer

The answer of A stone is thrown horizontally and follows...

Question 14

An object, tied to a string, moves in a circle at constant speed on a horizontal surface as shown. The direction of the displacement of this object, as it travels from W to X is: 
A) $\leftarrow$ 
B) $\downarrow$
C) $\uparrow$ 
D) 
E)

Accepted Answer

The answer of An object, tied to a string, moves...

Question 15

A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal component of 30 m/s. Using g = 10 m/s², the distance from launching to landing points is: A) 40 m B) 60 m C) 80 m D) 120 m E) 180 m

Accepted Answer

The time of flight can be calculated using the vertical component of the initial velocity and the acceleration due to gravity. The time to reach the maximum height is $t = \frac{v_{vertical}}{g} = \frac{20 m/s}{10 m/s^2} = 2 s$. Since the motion to the maximum height and back down is symmetrical, the total time of flight is $2t = 4 s$. The horizontal distance covered in this time, using the horizontal component of the velocity, is $distance = v_{horizontal} \times time = 30 m/s \times 4 s = 120 m$.

Question 16

A stone thrown from the top of a tall building follows a path that is:
A) circular
B) made of two straight line segments
C) hyperbolic
D) parabolic
E) a straight line

Accepted Answer

The path of a projectile, such as a stone thrown from a tall building, is best described by a parabolic shape due to the force of gravity acting on it. The stone will go up and then come down following a curved trajectory.

Question 17

An object is shot from the back of a railroad flatcar moving at 40 km/h on a straight horizontal road. The launcher is aimed upward, perpendicular to the bed of the flatcar. The object falls:
A) in front of the flatcar
B) behind the flatcar
C) on the flatcar
D) either behind or in front of the flatcar, depending on the initial speed of the object
E) to the side of the flatcar

Accepted Answer

The object maintains the horizontal velocity of the flatcar due to inertia, so when it is shot upward and falls back down, it does so in a path that keeps it above the flatcar, assuming no external horizontal forces act on it.

Question 18

Two projectiles are in flight at the same time. The acceleration of one relative to the other: A) is always 9.8 m/s² B) can be as large as 19.6 m/s² C) can be horizontal D) is zero E) none of these

Accepted Answer

Both projectiles experience the same acceleration due to gravity, so their relative acceleration is zero.

Question 19

A boy on the edge of a vertical cliff 20 m high throws a stone horizontally outwards with a speed of 20 m/s. It strikes the ground at what horizontal distance from the foot of the cliff? Use g = 10 m/s² A) 10 m B) 40 m C) 50 m D) 60 m E) none of these

Accepted Answer

The time taken for the stone to fall 20 m vertically can be calculated using the formula $ t = \sqrt{\frac{2h}{g}} $, where $ h = 20 $ m and $ g = 10 $ m/s². This gives $ t = \sqrt{\frac{2 \times 20}{10}} = 2 $ seconds. The horizontal distance is then $ \text{horizontal speed} \times \text{time} = 20 \, \text{m/s} \times 2 \, \text{s} = 40 \, \text{m} $.

Question 20

A large cannon is fired from ground level over level ground at an angle of 30 $\degree$ above the horizontal. The muzzle speed is 980 m/s. Neglecting air resistance, the projectile will travel what horizontal distance before striking the ground?
A) 4.3 km
B) 8.5 km
C) 43 km
D) 85 km
E) 170 km

Accepted Answer

The answer of A large cannon is fired from ground...

Quiz 4: Motion in Two and Three Dimensions

An airplane is flying north at 500 km/h. It makes a gradual 180 $\degree$ turn at constant speed, changing its direction of travel from north through east to south. The process takes 40 s. The average acceleration of the plane for this turn is:

A particle moves at constant speed in a circular path. The instantaneous velocity and instantaneous acceleration vectors are:

An airplane makes a gradual 90 $\degree$ turn while flying at a constant speed of 200 m/s. The process takes 20.0 seconds to complete. For this turn the magnitude of the average acceleration of the plane is:

A bullet shot horizontally from a gun:

A cannon fires a projectile as shown. The dashed line shows the trajectory in the absence of gravity; points MNOP correspond to the position of the projectile at one second intervals. If g = 10 m/s², the lengths X,Y,Z are:

A dart is thrown horizontally toward X at 20 m/s as shown. It hits Y 0.1 s later. The distance XY is:

The airplane shown is in level flight at an altitude of 0.50 km and a speed of 150 km/h. At what distance d should it release a heavy bomb to hit the target X? Take g = 10 m/s².

A toy racing car moves with constant speed around the circle shown below. When it is at point A its coordinates are x = 0, y = 3m and its velocity is (6 m/s) . When it is at point B its velocity and acceleration are:

A bomber flying in level flight with constant velocity drops a bomb before it is over the target. Neglecting air resistance, which one of the following is NOT true?

A stone is thrown outward from the top of a 59.4-m high cliff with an upward velocity component of 19.5 m/s. How long is the stone in the air? Assume that it lands on the ground below the cliff, and that the ground below the cliff is flat.

A stone is thrown horizontally and follows the path XYZ shown. The direction of the acceleration of the stone at point Y is: $A stone is thrown horizontally and follows the path XYZ shown. The direction of the acceleration of the stone at point Y is: A) \downarrow B) \rightarrow C) D) E)$

A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal component of 30 m/s. Using g = 10 m/s², the distance from launching to landing points is:

A stone thrown from the top of a tall building follows a path that is:

An object is shot from the back of a railroad flatcar moving at 40 km/h on a straight horizontal road. The launcher is aimed upward, perpendicular to the bed of the flatcar. The object falls:

Two projectiles are in flight at the same time. The acceleration of one relative to the other:

A boy on the edge of a vertical cliff 20 m high throws a stone horizontally outwards with a speed of 20 m/s. It strikes the ground at what horizontal distance from the foot of the cliff? Use g = 10 m/s²

A large cannon is fired from ground level over level ground at an angle of 30 $\degree$ above the horizontal. The muzzle speed is 980 m/s. Neglecting air resistance, the projectile will travel what horizontal distance before striking the ground?

Quiz 4: Motion in Two and Three Dimensions

An airplane is flying north at 500 km/h. It makes a gradual 180 °\degree° turn at constant speed, changing its direction of travel from north through east to south. The process takes 40 s. The average acceleration of the plane for this turn is:

A particle moves at constant speed in a circular path. The instantaneous velocity and instantaneous acceleration vectors are:

An airplane makes a gradual 90 °\degree° turn while flying at a constant speed of 200 m/s. The process takes 20.0 seconds to complete. For this turn the magnitude of the average acceleration of the plane is:

A bullet shot horizontally from a gun:

Which of the curves on the graph below best represents the vertical component vy versus t for a projectile fired at an angle of 45 °\degree° above the horizontal?

A cannon fires a projectile as shown. The dashed line shows the trajectory in the absence of gravity; points MNOP correspond to the position of the projectile at one second intervals. If g = 10 m/s2, the lengths X,Y,Z are:

A dart is thrown horizontally toward X at 20 m/s as shown. It hits Y 0.1 s later. The distance XY is:

The airplane shown is in level flight at an altitude of 0.50 km and a speed of 150 km/h. At what distance d should it release a heavy bomb to hit the target X? Take g = 10 m/s2.

A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45 °\degree° . With what speed was it thrown? Use g = 10 m/s2.