Ap1 Exam 1 Kinematics

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.

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The object moves the greatest distance during the time interval of 2 s and 3 s because the velocity is at its highest point during this interval. Since distance is equal to velocity multiplied by time, a higher velocity will result in a greater distance traveled within the same time interval. Therefore, the object moves the greatest distance during the time interval of 2 s and 3 s.

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.

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.

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.

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.

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.

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).

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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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.