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 deadheat (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

THERE IS YOURS SOLUTION

★ HENCE PROVED ★

HOPE IT HELPS

Answer 2

Hello Dear.

To Prove ⇒ 2(v₁ - v₂)(v₁a₂ - a₁v₂)/(a₂ - a₁)² 

[There is the little typing error in the Question. We can prove the above written Equation. It is correct.]

Proof ⇒

Let the distance covered by the car be x. & the time taken by the car be t.
[since, they reach the finishing point at the same time).

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

Using second equation of the motion,
   S = ut + 1/2 at²

where, S = Distance covered by the car, u = initial velocity, a = acceleration, and t = time taken.

∴ x = v₁ × t + 1/2 × a₁ × t²

 ⇒ x = v₁t + a₁t²/2

_________________________


For Second Car,
Acceleration = a₂
Initial Velocity = v₂
Time taken = t

Again using second equation of motion,

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

Now we have,
x = x

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 this value in the Equation distance covered by the First car.

∴ 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. :-)