Question

A student stands in an elevator and records his acceleration as a function of time. The data are shown in the graph above. At time t = 0, the elevator is at displacement x = 0 with velocity v = 0. Assume that the positive directions for displacement, velocity, and acceleration are upward. 

a. Determine the velocity v of the elevator at the end of each 5-second interval. 

i. Indicate your results by completing the following table. 

Time Interval (s) 0–5 5–10 10–15 15–20

ii. Plot the velocity as a function of time on the following graph.

b. Determine the displacement x of the elevator above the starting point at the end of each 5-second interval. 

i. Indicate your results by completing the following table. 

Time Interval (s) 0–5 5–10 10–15 15–20 

 x (m) _____ _____ _____ _____ 

ii. Plot the displacement as a function of time on the following graph.

Answer 1

a. i. Use the kinematic equation applicable  for constant acceleration: v = v₀ + at. For each time interval, substitute the initial velocity for that interval, the appropriate acceleration from the graph and a time of 5 seconds.

5 seconds: v = 0 + (0)(5s) = 0

10 seconds: v = 0 + (4m/s²)(5s) = 20mls 

15 seconds: v = 20mls + (0)(5s) = 20mls)

20 seconds: v = 20mls + (-4m/s²)(5s) = 0

b. i. Use the kinematic equation applicable for constant acceleration,  x = x₀ + v₀t + 1/2at². For each time interval, substitute the initial position for that interval, the initial velocity for that interval from part (a), the appropriate acceleration, and a time of 5 seconds.