Question

In a car race, car A takes a time t less than car B at the finish and passes the finishing point with speed v more than that of car B. Assuming that both the cars start from rest and travel with constant acceleration a₁ and a₂ repectively, the value of v interms of a₁ , a₂ and t is

(1) \((\frac{a_1a_2}{a_1+a_2})t\)

(2) (a₁+a₂)t

(3) \((\frac{a_1+a_2}{2})t\)

(4) \((\sqrt{a_1a_2})t\)

Answer 1

 (4) \((\sqrt{a_1a_2})t\)

Let A take t₁ seconds to finish the race. Then, according to the data given, B will take (t₁ + t) seconds to finish the race. Also, let v₁ be the velocity of B at the finishing point. Then the velocity of A at the finishing point will be (v₁ + v). Therefore, for A,

Now, total distance travelled by both A and B is same. i.e. S_A = S_B . Using the formula, S = ut + 1/2 at² , we get;

Subsituting value of t₁ from Eqn (4) in Enq (3) we have