Question

Two drag cars race. They line up at the starting line at rest. The winning car accelerates at a constant rate a and reaches the finish line with a final velocity v. The losing car accelerates at a constant rate a/4. How far had the losing car traveled, d, when the winning car crossed the finish line?

Answer 1

Answer:$d=\frac{v^2}{8a}$

Explanation:

Given

winning car accelerates with a and its final velocity is v

considering they both start from rest

time taken by winning car is

v=u+at

where u=initial velocity

a=acceleration

t=time

$v=at$

$t=\frac{v}{a}$

Now loosing car is accelerating with $\frac{a}{4}$

Distance traveled by loosing car in time t

$s_1=ut+\frac{at^2}{2\cdot 4}$

$s_1=0+\frac{a}{8}\times (\frac{v}{a})^2$

$s_1=\frac{v^2}{8a}$

Thus distance d traveled by loosing car is given by $d=\frac{v^2}{8a}$