I am going to do this question with actual numbers
let u0 = 20 m/s
let v0 = 28 m/s
let á = 5 m/s^2
let â = 4 m/s^2
case 1: á = 5 m/s^2
then u0 = 5t+ c
when t = 0 , 20 = 0 + c
velocity1 = 5t + 20
similarly velocity2 = 4t + 28
distance1 = (5/2)t^2 + 20t + k, but k = 0 since at t=0 distance = 0
distance1 = (5/2)t^2 + 20t
distance2 = 2t^2 + 28t
they went the same distance, so
2t^2 + 28t = (5/2)t^2 + 20t
(-1/2)t^2 + 8t = 0
t^2 - 16t = 0
t(t - 16) = 0
t = 0 or t = 16
either distance = 2t^2 + 28t (or the other one)
= 960 metres
how does that compare with your
s=(2(u_0-v_0)(u_0 â-v_0 á))/(â-á)^2
= (2(20-28)( 20*4 - 28*5))/(5-4)^2
= 2(-8)(-60)/1
= 960
Ok, it works, so now carefully repeat my steps using those variables.
Two cars start off a race with initial velocities u0 and v0, and travel in a straight line with uniform acceleration á and â. If the race ends in a dead beat, prove that the length s of the race is: s=(2(u_0-v_0)(u_0 â-v_0 á))/(â-á)^2
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