Question

In a car race on straight road, car A takes a times t less than car B at the finish and passes finishing point with a speed 'v' more than that of car B. Both the car start from rest and travel with constant acceleration a₁ and a₂ respectively. Then 'v' is equal to

A a₁+a₂/2t
B √2a₁a₂t
C 2a₁a₂/a1+a₂t
D √a₁a₂t

Answer 1

v is equal to √a₁a₂t. Hence the correct option is d.

The speed (v) of a car at the finish line in a straight road race is determined by the relationship between the accelerations (
`a_1` and
`a_2`), the time taken (t), and the initial speeds of the cars, assuming they start from rest. The correct expression is
`v=√(2a_1a_2t)`.

This equation is derived from the kinematic equation v^2 =u^2 +2as, where v is the final velocity, u is the initial velocity (which is zero in this case since both cars start from rest), a is the acceleration, and s is the displacement. Rearranging the equation to solve for v, we get v= 2as​.This equation represents the speed of the car at the finish line, with
`a_1`and
`a_2` being the constant accelerations of cars A and B, respectively, and t being the time taken. Hence the correct option is d.