Question

In a car race on straight road, car A takes a time ‘t’ less than car B at the finish and passes finishing point with a speed ‘v’ more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then ‘v’ is equal to
(A) [(2a₁a₂)/(a₁ + a₂)] t

(B) √(2a₁a₂) t
(C) √(a₁a₂) t (D) [(a₁ + a₂)/2] t

Answer 1

Thus the "v" is equal to V = √ a1a2t

Option (C) is correct.

Explanation:

Using 2nd equation of motion - time equation -

S = ut + 1 / 2 at^2

• S ---->  Displacement
• u  ---> Initial velocity
• a --> acceleration
• t ---> time

V(A) = ato

V(B) = (a1 - a2) to - ta2t

V(A) - V(B) = V = ( a1 - a2)to - a2t     ------(1)

XA = XB = 1/2 a1to^2 = 1/2 a2(to+t)^2

√ a1to = √ a2 (to + t)

√(√ a1 - √ a2)to = √ a2t    -----(2)

From equation (1) and (2)

V = (a1 - a2) √ a2t / √ a1 -√ a2 - a2t

(√ a1 + √ a2) √ a2t - a2t

V = √ a1a2t

Thus the "v" is equal to V = √ a1a2t