Answer
In 120 ms, the fist moves a distance of $~~49.75~cm$
Work Step by Step
On the graph, we can see that the the speed of the fist is a maximum when $t = 120~ms$
To find the distance that the fist moves, we can find the area under the velocity versus time curve for the fist. In part (a), we found that the area under the curve for $t=0$ to $t=50~ms$ is $13~cm$
We can add this to the area under the curve for $t=50~ms$ to $t=120~ms$
We can divide this area into four parts, including a triangle ($0.050~s$ to $0.090~s$), a rectangle ($0.050~s$ to $0.090~s$), a triangle ($0.090~s$ to $0.120~s$), and a rectangle ($0.090~s$ to $0.120~s$)
We can calculate each area separately:
$A_1 = \frac{1}{2}(0.040~s)(1.0~m/s) = 0.020~m$
$A_2 = (0.040~s)(4.0~m/s) = 0.160~m$
$A_3 = \frac{1}{2}(0.030~s)(2.5~m/s) = 0.0375~m$
$A_4 = (0.030~s)(5.0~m/s) = 0.150~m$
We can find the total area for $t=50~ms$ to $t=120~ms$:
$A = (0.020~m)+(0.160~m)+(0.0375~m)+(0.150~m)$
$A = 0.3675~m$
$A = 36.75~cm$
Then the area for $t=0~ms$ to $t=120~ms$ is:
$36.75~cm+13~cm = 49.75~cm$
In 120 ms, the fist moves a distance of $~~49.75~cm$
