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

Answer 1

Your Question is incomplete. You have exactly not written what we have to prove. As per as my experience and research on that topic, we have to prove that ⇒

Length of the course = 2(v₁ - v₂)(v₁a₂ - a₁v₂)/(a₂ - a₁)² 

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.

Now, According to the second equation of the motion,

S = ut + 1/2 at²
where, S = Distance covered by the car,
u = initial velocity,
a = acceleration, &
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

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

Now,
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

Answer 2

Let distance covered by cars = L and time taken is t

For first car :- initial velocity = v₁
Acceleration = a₁
Time taken = t
∴ use formula S = ut + 1/2at²
L = v₁t + 1/2 a₁t² --------(1)

For 2nd car :- initial velocity = v₂
acceleration = a₂
time taken = t
so, L = v₂t + 1/2 a₂t² -----(2)

From equations (1) and (2),
v₁t + 1/2a₁t² = v₂t + 1/2a₂t²
⇒(v₁ - v₂)t = 1/2(a₂ - a₁)t²
⇒t = 2(v₁ - v₂)/(a₂ - a₁) , put it in equation (1)

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