Answer
At $t=2.80~s$, the position of the center of block C is $~~x = 0.18~m$
Work Step by Step
In part (a), we found that between $t=0$ and $t = 0.80~s$, the velocity of block C is $0.60~m/s$
In part (a), we found that between $t=0.80~s$ and $t = 2.80~s$, the velocity of block C is $-0.15~m/s$
We can find the position of the center of block C at $t=2.80~s$:
$x = (0.60~m/s)(0.80~s)+(-0.15~m/s)(2.0~s)$
$x = 0.48~m-0.30~m$
$x = 0.18~m$
At $t=2.80~s$, the position of the center of block C is $~~x = 0.18~m$
