Answer
The vehicle travels a distance of $~~300~m~~$ from start to stop.
Work Step by Step
The vehicle accelerates for 10 seconds until it reaches a speed of $20~m/s$
We can find the position $x_1$ at the end of the acceleration period:
$x_1-x_0 = v_0~t+\frac{1}{2}at^2$
$x_1-0 = (0)(10~s)+\frac{1}{2}(2.0~m/s^2)(10~s)^2$
$x_1 = 100~m$
The vehicle decelerates for 20 seconds from a velocity of $20~m/s$ until it comes to rest.
We can find the final position $x_2$ at the end of the deceleration period:
$x_2-x_1 = v_0~t+\frac{1}{2}at^2$
$x_2-(100~m) = (20~m/s)(20~s)+\frac{1}{2}(-1.0~m/s^2)(20~s)^2$
$x_2 = 100~m+400~m-200~m = 300~m$
The vehicle travels a distance of $~~300~m~~$ from start to stop.
