Figure is a graph of the coordinate of a spider crawling along the x-axis. (a) Fraph tis velocity and acceleration as functionsof time. (b) In a motion diagram, show the position, velocity, and acceleration of the spider at the five times: `t=2.5 s`, t=10 s, t=20 s, t=30 s`,
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Explanation

a. `v_(x)` is the slope of the `x` versus `t` curve and `a_(x)` is the slope of the `v_(x)` versus `t` curve.
`t=O` to `t=5 s`: `x` versus `t` is a parabola, so `a_(x)` versus `t` is The curvature s positive so `a_(x)` is positive `v_(x)` versus `t` is a straight line with positive slope. `v_(x)=0`.
`t=5 s` to `t=25 s`: `x` versus `t` is a straight line, so `v_(x)` is constant and `0`. `a_(x)=0`. The slope of `x` versus `t` is positive, so `v_(x)` is positive.
`t=15 s` to `t=25 s`: x versus `t` is a parabola with negative curvature, so `a_(x)` is constant and negative, `v_(x)` versus `t` is a straight line with negative slope. The velocity is zero at `20 s`, positive to `15 s` to `20 s`, and negative to `20 s` to `25 s`. `t=25 s` to `t=35 s`: `x` versus `t` is a straight line, so `v_(x)` is constant and `a_(x)=0`. The slope of `x` versus `t` is negative, so `v_(x)` is negative.
`t=35 s` to `t=40 s`: `x` versus `t` is a parabola with positive curvature, so `a_(x)` is constant and positive.
`v_(x)` versus `t` is a straight line with positive slope. The velpcoty reaches zero at `t=40 s`
The graphs of `v_(x)(t)` are sketched in
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b. The motions diagrams are skerched in
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The spider speeds up for the first `5 s`, since `v_(x)` and `a_(x)` are both positive. Starting at `t=5 s`, the spider starts to slow down stops momentarily at `t=20 s`, and then moves in the opposite direction. At `t=35 s`, the spider sarts in the opppsits direction .At `t=35 s,` the spider starts to slow down again and stops at `t=40`.