Question

Question #d67f9

Answer 1

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`