Question

in a carrace car a takes 4z less than car b at the finish and passes the finishing point with a velocity v more thsn car b

Answer 1

For A

acceleration=a1

total time=t'-t

final velocity=u+v

For B

acceleration=a2

total time=t'

final velocity=u

s=(1/2)a1(t'-t)^2=(1/2)a2(t')^2

therfor

1)(1-t/t')=(a2/a1)^(1/2)

also

(u+v)^2=2a1s

(u)^2=2a2s

dividing both we get

2)(1+v/u)=(a1/a2)^(1/2)

but u=a2t'

so substitute this value in 2 and put value of t' from 1

therefore we get

v=t(a1a2)^(1/2)

v=t root of a1 a2