Question

Dr Alex’s cat, Rufus, walks along the top of a fence. As an observant scientist, she made measurements of the cat’s motion and constructed a graph of the cat’s velocity as a function of time. The fence top is straight and horizontal, and Dr Alex chose the positive x-direction to be to the right.

(cm/s) t (s) 0 1 2 3456

1. What is the cat’s acceleration at t=3.0 s and t=6.0 s.

2. Find the cat’s velocity at t=4.0 s and t=7.0 s.

3. Sketch the acceleration on a graph.

4. If the cat started at x0=0.0 m, where is it located at t=6.0 s?

Answer 1

Acceleration = slope of v- t graph At 6 sec v = 0 At 3 se v = 4 = 0 - 4/ 6 - 3 = -4/3 = -1.33 cm/s^2 v(4) = v(4) - v(3)/ 4 - 3 = -1.33 v(4) - 4/v(4) = -1.33 v(4) = 4 - 1.33 = 2.67 cm/s v(7) = v(7) - v(6)/7 - 6 = -1.33 = v(7) - 0 = -1.33 v(7) = -1.33 cm/s u_x = 8cm/s v_x = 0 a = -1.33 T 6 sec using v^2- u^2 = 2as s = (0)^2 - (8)^2/-2 times 1.33 implies 64/ 2.66 implies 24.06 cm