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Answer 1 · 2017-02-20T17:23:48

Let time taken to finish the race by car $A and B$ be $= t_A and t_B$ respectively.

We need to assume that cars start from rest and have respective acceleration $= a_1 and a_2$ .

Using the kinematic equation
$s=ut+1/2at^2$ we get
Race track $s =1/2 a_1t_A^2$
Also $s=1/2 a_2t_B^2$

Equating the length of race track in case of both cars
$a_1t_A^2 = a_2t_B^2$ ......… (1)

Also given is
$t_B= t_A+ t$ ........... (2)

Using the kinematic equation
$v=u+at$ we get velocity at the end of race for
car $A = a_1t_A and$ for car $B = a_2 t_B$

Given is
$a_1t_A = a_2t_B + v$ .....(3)

From this equation we need to eliminate $t_A and t_B$ to get the desired expression. (3) can be rewritten as
$v= a_1t_A-a_2t_B$

Substituting value of $a_1$ from (1)
$v= (a_2t_B^2)/t_A- a_2t_B$
$=>v= a_2(t_B/t_A)(t_B –t_A)$

Using (2) we get
$v= a_2 t (t_B/t_A)$

Using (1) we get
$v= a_2 tsqrt(a_1/a_2)$

Squaring both sides we get
$v^2= t^2a_1a_2$