Answer
In 50 ms, the fist moves a distance of $~~13~cm$
Work Step by Step
To find the distance that the fist moves, we can find the area under the velocity versus time curve for the fist.
We can divide this area into three parts, including a triangle ($0~s$ to $0.010~s$), a rectangle ($0.010~s$ to $0.050~s$), and a triangle ($0.010~s$ to $0.050~s$)
We can calculate each area separately:
$A_1 = \frac{1}{2}(0.010~s)(2.0~m/s) = 0.010~m$
$A_2 = (0.040~s)(2.0~m/s) = 0.080~m$
$A_3 = \frac{1}{2}(0.040~s)(2.0~m/s) = 0.040~m$
We can find the total area:
$A = (0.010~m)+(0.080~m)+(0.040~m)$
$A = 0.13~m$
$A = 13~cm$
In 50 ms, the fist moves a distance of $~~13~cm$.
