AP Physics 1 Exam Question

A cart is moving along a straight line. The position-time graph for the cart's motion is shown below:

a) Describe the motion of the cart during the first 10 seconds. Include information about the cart's speed, velocity, and acceleration during this time interval.

b) Determine the cart's average speed and average velocity from t = 10 s to t = 20 s.

c) Sketch the corresponding velocity-time graph for the cart's motion.

Answer

a) To describe the motion of the cart during the first 10 seconds, we can analyze the position-time graph.

From 0 to 5 seconds, the cart is at a constant position (x = 0), indicating it is not moving. Therefore, during this time interval, its speed is 0, velocity is 0, and acceleration is also 0.

From 5 to 10 seconds, the cart is moving in the positive x-direction at a constant velocity. The slope of the line joining these points on the position-time graph gives the velocity of the cart. The slope is positive, indicating the cart is moving in the positive direction. To calculate the velocity, we use the formula:

Here, the change in position is given by: Δx = xf - xi = 6 m - 0 m = 6 m.

The change in time is: Δt = tf - ti = 10 s - 5 s = 5 s.

Therefore, the velocity is: [ \text{Velocity} = \frac{6 , \text{m}}{5 , \text{s}} = 1.2 , \text{m/s} ]

Since the velocity is constant, the speed is also 1.2 m/s. The acceleration during this time interval is 0 since the velocity is constant.

b) To find the cart's average speed and average velocity from t = 10 s to t = 20 s, we need to calculate the total distance traveled and displacement during this time interval.

The total distance traveled is given by the area under the position-time graph. From t = 10 s to t = 20 s, the shape of the graph corresponds to a rectangle with dimensions 10 s (width) and 2 m (height). Therefore, the total distance traveled is:

The displacement is given by the change in position during this time interval. From t = 10 s to t = 20 s, the change in position is given by the difference in x-coordinates: Δx = xf - xi = 6 m - 2 m = 4 m.

Therefore, the average velocity is: [ \text{Average velocity} = \frac{\text{change in position}}{\text{change in time}} = \frac{4 , \text{m}}{10 , \text{s}} = 0.4 , \text{m/s} ]

The average speed is the total distance traveled divided by the time interval: [ \text{Average speed} = \frac{\text{total distance}}{\text{time interval}} = \frac{20 , \text{m}}{10 , \text{s}} = 2 , \text{m/s} ]

c) To sketch the corresponding velocity-time graph for the cart's motion, we need to determine the velocity of the cart at different time intervals.

• From t = 0 s to t = 5 s, the velocity is 0 m/s.
• From t = 5 s to t = 10 s, the velocity is constant at 1.2 m/s.
• From t = 10 s to t = 20 s, the velocity is constant at 0.4 m/s.

The resulting velocity-time graph would look like this:

The velocity-time graph is a horizontal line at 1.2 m/s from t = 5 s to t = 10 s, and then a horizontal line at 0.4 m/s from t = 10 s to t = 20 s.