Question

A car acceleration from rest at a constant rate 10 $m/s^2$ for some time after which it accelerates at constant rate 5 $m/s^2$ and then comes to rest. Total time elapsed is 15 s. Maximum velocity of the car is
(i) while average velocity of the car during its deceleration is
(ii) Distance traveled by the car at the instant when its velocity is half of its maximum value is (iii) or (iv).

Answer 1

Total time to travel in $6$ sec

acceleration $=2m/s^2$

retardation $=4m/s^2$

$\therefore$ $v=u+at$ $s=ut+\frac{1}{2}a{t}^2$

$v_1=2t$

$v_f=2t-4(6-t)=0$

$=2t-2u+4t=0$

$6t=24$

$t=4$

Now $S=S_1+S_2$

$=(0\times 4+\frac{1}{2}\times 2\times 4^2)+(8\times 2-\frac{1}{2}\times 4\times 2^2)$

$=16+(16-8)$

$S=24$