Question

A block is suspended from a spring, pulled down, and released. The block’s position-versus-time graph is shown in FIGURE P2.38. a. At what times is the velocity zero? At what times is the velocity most positive? Most negative? b. Draw a reasonable velocity-versus-time graph.

Short answer

Part (a)

The velocity is zero at $t=0s,1s,2s,...$

The velocity is most positive at$t=0.5s,2.5s,...$

The velocity is most negative at $t=1.5s,3.5s,...$

Part (b)

The velocity-time graph is

Step-by-step solution

Given information

The position versus time graph of a block is shown as

Part (a)

The velocity is zero when the slope of the position-time graph is zero, it is most positive when the slope is most positive and is most negative when the slope is most negative.

The slope is zero at $t=0s,1s,2s,...$

The slope is most positive at $t=0.5s,2.5s,...$

The slope is most negative at$t=1.5s,3.5s,...$

Draw velocity-versus-time graph.

With the help of the information in step 2, the velocity-versus-time graph is drawn as below