Answer
At $t=5.0~s$, the coordinate of the particle is $~~15.0~m$
Work Step by Step
To find the displacement of the particle, we can find the area under the velocity versus time curve between $t=0$ and $t = 5.0~s$.
We can divide this area into four parts, including a triangle ($0~s$ to $2.0~s$), a rectangle ($2.0~s$ to $4.0~s$), a triangle ($4.0~s$ to $5.0~s$), and a rectangle ($4.0~s$ to $5.0~s$)
We can calculate each area separately:
$A_1 = \frac{1}{2}(2.0~s)(4.0~m/s) = 4.0~m$
$A_2 = (2.0~s)(4.0~m/s) = 8.0~m$
$A_3 = \frac{1}{2}(1.0~s)(2.0~m/s) = 1.0~m$
$A_4 = (1.0~s)(2.0~m/s) = 2.0~m$
We can find the total area:
$A = (4.0~m)+(8.0~m)+(1.0~m)+(2.0~m)$
$A = 15.0~m$
At $t=5.0~s$, the coordinate of the particle is $~~15.0~m$.
