Question

Two cars start off to race with velocities v1 and v2 and travel in a straight line with uniform acceleration a1 and a2 respectively. If the race ends in a dead heat ( i.e they reach the finishing point at the same time) Prove the length of the course is

2(v1-v2)(v1a2-v2a2)/(a1-a2)^2

Answer 1

We have To Prove :- 2(v₁ - v₂)(v₁a₂ - a₁v₂)/(a₂ - a₁)² 


Now Proof :-

Let us the distance covered = x
Time taken by both the car = t 
[since, both the Car reached the finishing point at the same time),

So,
For the Ist Car,
Acceleration of the Ist car = a₁
Initial Velocity of the Ist car = v₁
Time taken by the Ist car to reach the Finishing point = t sec.

Using the 2nd equation of the motion,

S = ut + 1/2 at²
Here,
S :- Distance covered by the  car,
u :- initial velocity of Ist car,
a :- acceleration, (a₁) &
 t :- time taken

Therefore
x = v₁ × t + 1/2 × a₁ × t²
 x = v₁t + a₁t²/2

For 2nd Car

Acceleration of the 2nd car= a₂
Initial Velocity of the 2nd car = v₂
Time taken = t

Therefore
S = v₂ × t + 1/2 × a₂ × t²
 x = v₂ + a₂ t²/2

Now ,We know,
 x = x [Distance covered by both the car is equal]
v₁t + a₁t²/t = v₂t + a₂ t²/2
v₁t - v₂t = a₂t²/2 - a₁t²/2
 t(v₁ - v₂) = t²/2(a₂ - a₁)
 t = 2(v₁ - v₂) ÷ (a₂ - a₁)

Now, Putting  values in the Equation of the distance covered by the Ist car.
We get,

 x = v₁[2(v₁ - v₂)/(a₂ - a₁)] + 1/2 a₁ {2(v₁ - v₂)/(a₂ - a₁)}²  
 x = [(v₁ - v₂)/(a₂ - a₁)] {2v₁ + 2a₁(v₁ - v₂)/(a₂ - a₁)}  
 x = [(v₁ - v₂)/(a₂ - a₁)] {2v₁(a₂ - a₁) + 2a₁(v₁ - v₂)}/(a₂ - a₁)
 x = (v₁ - v₂)/(a₂ - a₁)² [2v₁a₂ - 2v₁a₁ + 2v₁a₁ -2a₁v₂]
  x = 2(v₁ - v₂)(v₁a₂ - a₁v₂)/(a₂ - a₁)² 


Hence Proved,


Hope it helps.