Answer
(a) At $t = 3~s$
$a = 0$
At $t = 7~s$
$a = 6.3~m/s^2$
At $t = 11~s$
$a = -11.3~m/s^2$
(b) From $t = 0$ to $t = 5~s$
The officer goes 100 meters.
From $t = 0$ to $t = 9~s$
The officer goes 230 meters.
From $t = 0$ to $t = 13~s$
The officer goes 320 meters.
Work Step by Step
(a) The instantaneous acceleration is the slope of the velocity-time graph at each time t.
At $t = 3~s$, the slope of the velocity-time graph is zero, so $a = 0$
At $t = 7~s$
$slope = \frac{45~m/s-20~m/s}{9~s-5~s} = \frac{25~m/s}{4~s} = 6.3~m/s^2$
The slope of the velocity-time graph is $6.3~m/s^2$, so $a = 6.3~m/s^2$
At $t = 11~s$
$slope = \frac{0-45~m/s}{13~s-9~s} = \frac{-45~m/s}{4~s} = -11.3~m/s^2$
The slope of the velocity-time graph is $-11.3~m/s^2$, so $a = -11.3~m/s^2$
(b) The displacement is the area under the velocity-time graph.
From $t = 0$ to $t = 5~s$
The area under the velocity-time graph is $(20~m/s)(5~s) = 100~m$
The officer goes 100 meters.
From $t = 0$ to $t = 9~s$
The area under the velocity-time graph is $(20~m/s)(9~s) +\frac{1}{2}(25~m/s)(4~s)= 230~m$
The officer goes 230 meters.
From $t = 0$ to $t = 13~s$
The area under the velocity-time graph is $(20~m/s)(9~s) +\frac{1}{2}(25~m/s)(4~s)+\frac{1}{2}(45~m/s)(4~s)= 320~m$
The officer goes 320 meters.
