Representing Motion AP  Physics 1 FRQ

Ans:

(a) On a velocity-versus-time graph, the displacement is the area between the graph and the x-axis. Area above the x-axis is positive displacement and area below the x-axis is negative displacement. The displacement from time t = 0 s to t = 6 s can be evaluated as
t = 0 s to t = 2 s: (2 s)(1 m/s) + \(\frac{1}{2}\)(2 s)(1 m/s) = 3 m
t = 2 s to t = 3 s: (1 s)(2 m/s) = 2 m
t = 3 s to t = 4 s: (1 s)(1 m/s) + \(\frac{1}{2}\)(1 s)(1 m/s) = 1.5 m
t = 4 s to t = 5 s: \(\frac{1}{2}\)(1 s)(1 m/s) = 0.5 m
t = 5 s to t = 6 s: \(\frac{1}{2}\)(1 s)(−1 m/s) = −0.5 m
The total displacement from time t = 2 s to t = 6 s is: 3 m + 2 m + 1.5 m + 0.5 m − 0.5 m = 6.5 m.

(b) The speed of the object is increasing when the magnitude of the velocity is increasing. This occurs from time t = 0 s to t = 2 s and t = 5 s to t = 6 s.

(c)

Section a tells you the object is speeding up in the positive direction (parabola with an increasingly steeper slope). Section b tells you the object is traveling with constant speed in the positive direction (linear line). Section c tells you the object is slowing down but still moving in the positive direction (parabola with a decreasingly steeper slope). Section d tells you the object is moving in the negative direction and speeding up (parabola with an increasingly steeper slope).