Question

An automobile and a truck start from rest at the same instant, with the automobile initially at some distance behind the truck. Then truck has a constant acceleration of 2.2 $m/s^2$ and the automobile has an acceleration of 3.5 $m/s^2$. The automobile overtakes the truck when it (truck) has moved 60 m.
(a) How much time does it take to automobile to overtake the truck?
(b) How far was the automobile behind the truck initially?
(c) What is the speed of each during overtaking?

Answer 1

$\left(a\right)$After $t$ sec the truck moved

$s=ut+\frac{1}{2}a{t}^2$

Initial velocity$u=0$

$60=\frac{1}{2}\times 2.20\times t^2$

$t^2=54.54$

$t=7.39sec$

$\left(b\right)$ Let the distance be $d$ meters

When they met, automobile moved

$60+d=\frac{1}{2}\times 3.50\times t^2$

$60+d=\frac{1}{2}\times 3.50\times {7.39}^2$

$60+d=95.57$

$d=95.57-60$

$d=35.57m$

$\left(c\right)$Velocity of truck

$v_T=u+at$

$v_T=2.20\times 7.39=16.258m/s$

Velocity of automobile

$v_A=u+at$

$v_A=3.50\times 7.39=25.865m/s$