Question #d280e

Apr 26, 2017

$1000 m$

Explanation:

Distance traveled $s$ in time $t$. Applicable kinematic equation is
$s = u t + \frac{1}{2} a {t}^{2}$
where $a$ is acceleration and $u$ is initial velocity.

Car a
${s}_{a} = 0 \times 60 + \frac{1}{2} \left(0.75\right) {\left(60\right)}^{2}$
${s}_{a} = 1350 m$

Car b
${s}_{b} = 0 \times 60 + \frac{1}{2} \left(0.25\right) {\left(60\right)}^{2}$
${s}_{a} = 450 m$

Difference is $= 1350 - 450 = 900 m$
When cars started Car a was ahead of Car b by$= 100 m$
After $1 \min$ Car a is ahead of Car b by$= 900 + 100 m = 1000 m$