Ap1 Exam 1 Kinematics
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The graph above shows the velocity versus time for an object moving in a straight line. For which time interval does the object move the greatest distance?
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A. 0 s and 1 s
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B. 1 s and 2 s
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C. 2 s and 3 s
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D. 3 s and 4 s
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About This Quiz
The 'AP1 Exam 1 Kinematics' quiz assesses understanding of motion in a straight line, effects of gravity on motion, and vector analysis through practical problems. It evaluates skills in interpreting motion graphs and calculating distances, velocities, and acceleration, crucial for students in physics.
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- 2.
The graph shows the velocity versus time for an object moving in a straight line. At what time after t = 0 s does the object again pass through its initial position?
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A. Between 0 s and 1 s
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B. Between 1 s and 2 s
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C. Between 2 s and 3 s
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D. Between 3 s and 4 s
Correct Answer
A. B. Between 1 s and 2 sExplanation
The graph shows that the velocity of the object is initially positive and then becomes negative. This indicates that the object is moving in one direction and then changes its direction. The object passes through its initial position when its velocity is zero, which occurs between 1 s and 2 s on the graph. Therefore, the correct answer is B. Between 1 s and 2 s.Rate this question:
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- 3.
A small cart is placed on a ramp that is curved as shown. When released from rest, which of the following best represent what happens to the magnitude of the cart’s velocity and acceleration as it moves with negligible friction down the ramp?
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A. Increasing velocity and increasing acceleration
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B. Decreasing velocity and increasing acceleration
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C. Increasing velocity and decreasing acceleration
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D. Increasing velocity and constant acceleration
Correct Answer
A. C. Increasing velocity and decreasing accelerationExplanation
As the cart moves down the ramp, its velocity increases because it is moving in the direction of the ramp's incline. However, the acceleration decreases because the gravitational force acting on the cart is balanced by the normal force from the ramp, resulting in a decrease in the net force and hence a decrease in acceleration. Therefore, option C, increasing velocity and decreasing acceleration, best represents what happens to the cart's velocity and acceleration.Rate this question:
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- 4.
A ball initially at rest is dropped from a height, y, above the floor, and it hits the floor after 1.5 s. From what height should the ball be dropped so that it takes 3.0 s to strike the floor?
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A. 0.7y
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B. 1.4y
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C. 2y
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D. 4y
Correct Answer
A. D. 4yExplanation
The time it takes for an object to fall from a certain height is directly proportional to the square root of the height. In this case, the ball takes 1.5 seconds to hit the floor when dropped from height y. Therefore, if the ball is dropped from a height that is four times higher (4y), it will take twice as long (3 seconds) to hit the floor.Rate this question:
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- 5.
A truck traveled north for 400 m at 5 m/s and then it traveled east for 300 m at 4.3 m/s. The magnitude of the average velocity of the truck was most nearly
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A. 1.2 m/s
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B. 3.3 m/s
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C. 4.6 m/s
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D. 6.6 m/s D. 6.6 m/s
Correct Answer
A. B. 3.3 m/sExplanation
The magnitude of average velocity is calculated by finding the total displacement and dividing it by the total time taken. In this case, the total displacement is the vector sum of the north and east displacements, which can be calculated using the Pythagorean theorem. The north displacement is 400 m and the east displacement is 300 m. Using the Pythagorean theorem, the total displacement is approximately 500 m. The total time taken can be calculated by adding the time taken for each displacement, which is 80 seconds. Therefore, the magnitude of average velocity is approximately 500 m / 80 s = 6.25 m/s. Since the question asks for the answer that is most nearly correct, the closest option is B. 3.3 m/s.Rate this question:
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- 6.
A basketball is thrown toward a hoop with an initial velocity of Vo with a trajectory as shown. Consider air friction to be negligible. Point R is the highest point on the path. Points Q and S are the same height above the ground.Which of the following correctly ranks the speeds of the basketball at the four locations?
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A. vP
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B. vS
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C. vQ = vS
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D. vR
Correct Answer
A. D. vRExplanation
At point R, the basketball reaches its highest point on the path. At this point, the basketball's velocity is zero as it momentarily stops before starting to fall back down. Therefore, the speed of the basketball at point R is the lowest among the four locations.Rate this question:
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- 7.
A basketball is thrown toward a hoop with an initial velocity of Vo with a trajectory as shown. Consider air friction to be negligible. Point R is the highest point on the path. Points Q and S are the same height above the ground.Which of the following diagrams best shows the direction of the acceleration of the ball at point Q ?
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A
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B
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C
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D
Correct Answer
A. DExplanation
At point Q, the ball is at its highest point on the path. At this point, the ball is moving upwards, which means its velocity is decreasing. According to Newton's second law of motion, the acceleration of an object is in the direction of the net force acting on it. Since the velocity is decreasing, the net force must be in the opposite direction of the velocity. Therefore, the acceleration at point Q is directed downwards, opposite to the direction of the velocity. Diagram D shows the acceleration of the ball at point Q in the correct direction.Rate this question:
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- 8.
Which of the following pairs of graphs best represents the vertical components of the velocity and acceleration, v and a, respectively, of the ball’s trajectory as functions of time t ?
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A
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B
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C
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D
Correct Answer
A. DExplanation
Graph D represents the vertical components of velocity and acceleration of the ball's trajectory as functions of time. The velocity graph shows a constant positive velocity, indicating that the ball is moving upwards at a constant speed. The acceleration graph shows a constant negative acceleration, indicating that the ball is experiencing a constant downward force due to gravity. This combination of a constant positive velocity and a constant negative acceleration is consistent with the ball's trajectory when it is thrown upwards and starts to fall back down.Rate this question:
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- 9.
The table shown represents the position of a laboratory cart at various instants in time as recorded by a motion detector. Which of the following best represents the velocity of the cart at time t = 0.5 s?
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A. 4 cm/s
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B. 7 cm/s
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C. 8 cm/s
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D. 28 cm/s
Correct Answer
A. C. 8 cm/s -
- 10.
A low friction cart is placed at the top of a ramp as shown. Upon release from rest, which of the following graphs best represents the velocity of the cart as a function of time?
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A
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B
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C
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D
Correct Answer
A. CExplanation
The correct answer is C because initially, when the cart is released from rest, it will experience an acceleration due to the force of gravity pulling it down the ramp. As time passes, the cart will continue to accelerate until it reaches a maximum velocity. After reaching the maximum velocity, the cart will maintain a constant velocity as it continues to move down the ramp. Therefore, the graph that best represents this scenario is graph C, which shows an initial increase in velocity followed by a constant velocity.Rate this question:
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- 11.
A water balloon is launched from ground level toward a building. As the water balloon flies through the air, it strikes the second floor of the building after passing the peak of its trajectory. Disregarding any effects of air resistance and using the launch point as the origin and up as the positive (+) direction, which of the following shows the correct signs regarding the balloon’s vertical position, vertical velocity, and vertical acceleration the instant before it strikes the building?
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A
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B
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C
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D
Correct Answer
A. C -
- 12.
A projectile is fired from the surface of Earth with a speed of 200 meters per second at an angle of 30°above the horizontal. If the ground is level, what is the maximum height reached by the projectile?
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A. 5 m
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B. 10 m
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C. 500 m
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D. 1000 m
Correct Answer
A. C. 500 mExplanation
When a projectile is fired at an angle above the horizontal, it follows a curved path called a parabola. The maximum height reached by the projectile occurs at the highest point of this parabolic path.
To find the maximum height, we can use the equation for the vertical motion of a projectile:
H = (v^2 * sin^2(θ)) / (2 * g)
Where:
H = maximum height
v = initial velocity of the projectile (200 m/s)
θ = launch angle (30 degrees)
g = acceleration due to gravity (9.8 m/s^2)
Plugging in the values, we get:
H = (200^2 * sin^2(30)) / (2 * 9.8)
H = (40000 * (1/4)) / 19.6
H = 10000 / 19.6
H ≈ 510.2 m
Therefore, the maximum height reached by the projectile is approximately 510.2 meters, which is closest to option C (500 m).Rate this question:
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- 13.
The graph provided shows the position, x, as a function of time, t. for a motorcycle moving in a straight line. Which of the following statements is the best description of the motion of the motorcycle?
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A. Slowing down, moving at a constant speed, then speeding up
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B. Speeding up, moving at a constant speed, then slowing down
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C. Moving with an increasing speed the entire time
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D. Climbing a steep hill and then reaching the top
Correct Answer
A. B. Speeding up, moving at a constant speed, then slowing downExplanation
Based on the graph, we can observe that the position of the motorcycle initially increases at a slow rate, indicating that it is speeding up. Then, the position remains constant for a period of time, indicating that it is moving at a constant speed. Finally, the position decreases at a slow rate, indicating that it is slowing down. Therefore, option B is the best description of the motion of the motorcycle.Rate this question:
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- 14.
A body moving in the positive x direction passes the origin at time t = 0. Between t = 0 and t = 1 s, the body has a constant speed of 24 m/s. At t = 1 s, the body is given a constant acceleration of 6 m/s2 in the negative x direction. The position x of the body at t = 11 s is
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A. –99 m
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B. –36 m
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C. 36 m
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D. 99 m
Correct Answer
A. B. –36 mExplanation
The body is initially moving in the positive x direction and has a constant speed of 24 m/s between t = 0 and t = 1 s. This means that in the first second, the body travels a distance of 24 meters in the positive x direction.
At t = 1 s, the body is given a constant acceleration of 6 m/s^2 in the negative x direction. This means that the body starts decelerating and its velocity decreases by 6 m/s every second.
Since the body is decelerating, it will eventually come to a stop and start moving in the negative x direction.
At t = 11 s, the body has been decelerating for 10 seconds. The change in velocity due to the deceleration is 6 m/s * 10 s = -60 m/s.
Since the body was initially moving in the positive x direction, the change in velocity of -60 m/s means that the body is now moving in the negative x direction with a velocity of -60 m/s.
The total distance traveled in the negative x direction is the product of the velocity and time, which is -60 m/s * 10 s = -600 m.
Since the body passed the origin at t = 0, the position x of the body at t = 11 s is -600 m + 0 m = -600 m.
Therefore, the correct answer is B. -36 m.Rate this question:
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- 15.
A ball is shot into the air with a positive initial velocity as shown. Consider air resistance as negligible. Which of the following launch angles will cause the ball to remain in the air the longest time?
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A. 35°
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B. 45°
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C. 55°
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D. 65°
Correct Answer
A. D. 65°Explanation
A launch angle of 65° will cause the ball to remain in the air the longest time because at this angle, the initial vertical velocity component is the smallest compared to the other angles. This means that the ball will spend more time in the air before reaching its maximum height and falling back down. The higher the launch angle, the higher the initial vertical velocity component, causing the ball to reach its maximum height faster and spend less time in the air. Therefore, a launch angle of 65° will result in the longest time in the air.Rate this question:
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- 16.
An object slides off a roof 10 m above the ground with an initial horizontal speed of 5 m/s as shown. If air resistance is negligible, what is the velocity of the object the instant before impact with the ground?
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A. 5 m/s vertically and 5 m/s horizontally
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B. 5 m/s vertically and 19 m/s horizontally
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C. 14 m/s vertically and 19 m/s horizontally
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A. 14 m/s vertically and 5 m/s horizontally
Correct Answer
A. A. 14 m/s vertically and 5 m/s horizontallyExplanation
The object is sliding off the roof with an initial horizontal speed of 5 m/s, and there is no air resistance. This means that the horizontal velocity remains constant throughout the motion. The vertical velocity, on the other hand, changes due to the effect of gravity. As the object falls, its vertical velocity increases by 9.8 m/s every second. When the object reaches the ground, its vertical velocity will be 14 m/s downward. Therefore, the velocity of the object just before impact is 14 m/s vertically and 5 m/s horizontally.Rate this question:
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- 17.
An object is dropped from rest from the top of a 400 m cliff on Earth. If air resistance is negligible, what is the distance the object travels during the first 6 s of its fall?
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A. 30 m
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B. 60 m
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C. 120 m
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D. 180 m
Correct Answer
A. D. 180 mExplanation
The distance an object travels during free fall can be determined using the equation d = 1/2gt^2, where d is the distance, g is the acceleration due to gravity, and t is the time. In this case, the object is dropped from rest, so its initial velocity is 0. Plugging in the values, we get d = 1/2 * 9.8 m/s^2 * (6 s)^2 = 1/2 * 9.8 m/s^2 * 36 s^2 = 176.4 m. Rounding to the nearest whole number, the object travels approximately 180 m during the first 6 s of its fall.Rate this question:
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- 18.
The position of an object is given by the equationx = 3.0t2 + 1.5t + 4.5where x is in meters and t is in seconds. What is the instantaneous acceleration of the object at t =3.0 s?
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A. 3.0 m/s^2
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B. 6.0 m/s^2
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C. 9.0 m/s^2
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D. 19.5 m/s^2
Correct Answer
A. B. 6.0 m/s^2Explanation
The instantaneous acceleration of an object can be found by taking the second derivative of the position equation with respect to time. The first derivative of the position equation is the velocity equation, which is given by v = 6.0t + 1.5. Taking the derivative of the velocity equation with respect to time gives the acceleration equation, which is a = 6.0 m/s^2. Therefore, the instantaneous acceleration of the object at t = 3.0 s is 6.0 m/s^2.Rate this question:
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- 19.
A gazelle is standing still, when it notices a nearby lion. The gazelle accelerates uniformly in a straight line away from the lion. Which of the following pairs of graphs shows the distance traveled versus time and the speed versus time for the gazelle after it notices the lion?
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A
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B
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C
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D
Correct Answer
A. CExplanation
Option C shows the distance traveled versus time graph as a straight line with positive slope, indicating that the distance increases with time. This is because the gazelle is accelerating uniformly away from the lion. Option C also shows the speed versus time graph as a straight line with positive slope, indicating that the speed of the gazelle is increasing at a constant rate. This is consistent with the gazelle accelerating uniformly.Rate this question:
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- 20.
A track star in the long jump goes into the jump at 12 m/s and launches herself at 20.0° above the horizontal. What is the magnitude of her horizontal displacement?
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A. 4.6 m
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B. 13 m
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C. 9.2 m
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D. 15 m
Correct Answer
A. C. 9.2 mExplanation
The magnitude of the horizontal displacement can be found using the formula: horizontal displacement = initial velocity * time * cos(angle). In this case, the initial velocity is 12 m/s and the angle is 20.0°. Since the time is not given, we can assume that it is equal to 1 second for simplicity. Therefore, the horizontal displacement is equal to 12 m/s * 1 s * cos(20.0°) = 9.2 m.Rate this question:
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- 21.
In a coordinate system, the magnitude of the x component of a vector and q, the angle between the vector and x-axis, are known. The magnitude of the vector equals the x component
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A. divided by the cosine of theta.
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B. divided by the sine of theta.
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C. multiplied by the cosine of theta.
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D. multiplied by the sine of theta.
Correct Answer
A. A. divided by the cosine of theta.Explanation
The magnitude of a vector can be calculated using the formula: magnitude = x component / cosine(theta). This formula is derived from the definition of cosine as the ratio of the adjacent side (x component) to the hypotenuse (magnitude of the vector) in a right triangle formed by the vector and the x-axis. Therefore, the correct answer is A. divided by the cosine of theta.Rate this question:
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- 22.
A teacher has challenged a student to hit a target with a dart gun. The dart is launched from ground level and the target is placed one meter above the ground. For the first attempt, the student launches the dart at an angle of 45° but the dart goes too far. Which of the following options could the student pick to produce a more accurate second shot? Select two answers.CHOOSE TWO
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A. Increase the launch angle
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B. Decrease the launch angle
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C. Increase the launch height
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D. Decrease the target height
Correct Answer(s)
A. A. Increase the launch angle
A. B. Decrease the launch angleExplanation
By increasing the launch angle, the student can make the dart go higher and have a shorter horizontal distance, which can help in hitting the target accurately. By decreasing the launch angle, the student can make the dart have a longer horizontal distance, which can also help in hitting the target accurately. Increasing the launch height or decreasing the target height will not directly affect the accuracy of the shot.Rate this question:
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- 23.
The velocity vs. time graph for an elevator is shown. For the time interval 0 s to 4 s, which of the following best describes the motion of the elevator? Select two answers.
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A) The elevator moves upward with decreasing velocity.
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B) The elevator moves downward with increasing velocity.
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C) The elevator returns to its starting position.
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D) The elevator moves with constant acceleration.
Correct Answer(s)
A. C) The elevator returns to its starting position.
A. D) The elevator moves with constant acceleration.Explanation
The velocity vs. time graph shows that the elevator starts at a positive velocity and gradually decreases until it reaches zero at around 2 seconds. After that, the velocity becomes negative, indicating that the elevator is moving downward. The fact that the velocity remains constant during this downward motion suggests that the elevator is moving with constant acceleration. Finally, at around 4 seconds, the velocity reaches zero again, indicating that the elevator has returned to its starting position. Therefore, the correct answers are C) The elevator returns to its starting position and D) The elevator moves with constant acceleration.Rate this question:
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- 24.
A student gives a cart a push up a ramp. The resulting position vs. time graph is displayed. At t = 1 second, which of the following are true? Select two answers.
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A. The displacement is zero.
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B. The velocity is zero.
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C. The acceleration is zero.
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D. The cart changes directions.
Correct Answer(s)
A. B. The velocity is zero.
A. D. The cart changes directions.Explanation
At t = 1 second, the position vs. time graph shows that the cart has reached its maximum position and starts moving back down the ramp. This means that the cart changes directions at t = 1 second. Additionally, since the cart is momentarily at rest at the maximum position, the velocity is zero at t = 1 second. Therefore, the correct answers are B. The velocity is zero and D. The cart changes directions.Rate this question:
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- 25.
A target T lies flat on the ground 3 m from the side of a building that is 10 m tall, as shown. A student rolls a ball off the horizontal roof of the building in the direction of the target, but misses. Consider air resistance as negligible. Which of the following can be determined from the information given? Select two answers.
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A. The speed of the ball as it strikes the ground
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B. The time the ball is in the air
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C. The acceleration of the ball 1 s before it strikes the ground
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D. The horizontal speed of the ball
Correct Answer(s)
A. B. The time the ball is in the air
A. C. The acceleration of the ball 1 s before it strikes the groundExplanation
From the given information, we know the height of the building (10 m) and the distance between the building and the target (3 m). Since air resistance is neglected, we can assume that the ball will follow a projectile motion. Using the equations of motion, we can determine the time the ball is in the air (Answer B) by using the equation for vertical motion. We can also determine the acceleration of the ball 1 second before it strikes the ground (Answer C) by using the equation for vertical motion and assuming that the only force acting on the ball is gravity. The speed of the ball as it strikes the ground (Answer A) and the horizontal speed of the ball (Answer D) cannot be determined from the given information.Rate this question:
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Mar 21, 2023Quiz Edited by
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Sep 02, 2015Quiz Created by
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