A student drives a moped along a straight road as described by the velocity versus time graph in the figure. The divisions along the horizontal axis represent 1.5 s and the divisions along the vertical axis represent 4.5 m/s. Sketch this graph in the middle of a sheet of graph paper. (Do this on paper. Your instructor may ask you to turn in your work.)

(c) What is the acceleration at t = 9.0 s?

(d) Find the position (relative to the starting point) at t = 9.0 s.

(e) What is the moped's final position at t = 13.5 s?

To answer these questions, we need to analyze the velocity versus time graph of the moped. Let's start by finding the acceleration at t = 9.0 s.

(a) To determine the acceleration at a specific time, we need to look at the slope of the velocity versus time graph at that point. Recall that the slope of a graph represents the rate of change, which in this case is acceleration.

(b) To find the acceleration at t = 9.0 s, draw a tangent line to the graph at that time point. Then, determine the slope of the tangent line. The slope can be determined by finding the change in velocity divided by the change in time over a small interval.

(c) Once you have determined the slope, convert it to actual units using the given scale. In this case, the divisions along the vertical axis represent 4.5 m/s.

(d) Now, let's move on to finding the position (relative to the starting point) at t = 9.0 s.

(e) Position can be determined by integrating the velocity with respect to time. In other words, we need to find the area under the velocity versus time graph from the starting point to t = 9.0 s.

(f) Draw a vertical line from t = 9.0 s on the horizontal axis to the velocity versus time graph. Then, draw a horizontal line from the top of this vertical line to the velocity axis. This forms a triangle, and the area of this triangle represents the displacement.

(g) Calculate the area of the triangle using the formula for the area of a triangle: A = (base * height) / 2. In this case, the base is the time of 9.0 s, and the height is the velocity at t = 9.0 s.

(h) Finally, multiply the displacement by (-1) if the moped's motion is in the negative direction, indicating that it is moving backwards.

(i) Moving on to finding the moped's final position at t = 13.5 s.

(j) Similarly to finding the position at t = 9.0 s, we need to integrate the velocity with respect to time from the starting point to t = 13.5 s.

(k) Repeat the process of drawing a vertical line from t = 13.5 s on the horizontal axis to the velocity versus time graph, and draw a horizontal line from the top of this vertical line to the velocity axis.

(l) Determine the area of the resulting shape, which may involve finding multiple areas of triangles and rectangles depending on the graph.

(m) Just like before, multiply the displacement by (-1) if the moped's motion is backwards.

Now that we have gone through the steps, you can follow them to find the acceleration at t = 9.0 s, the position at t = 9.0 s, and the final position at t = 13.5 s based on the given velocity versus time graph. Remember to follow the instructions on sketching the graph on a sheet of graph paper as well.