In a car race on straight road, car A takes a time less than car B at the finish and passes finishing point with a speed more than of car B. Both the cars start from rest and travel with constant acceleration and respectively. Then is equal to:

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 of car B. Both the cars start from rest and travel with constant acceleration $a_{1}$ and $a_{2}$ respectively. Then $' v'$ is equal to:

Smart Achievers · Accepted Answer

(3)

Let car B take time T. Then car A takes $T - t$ .

Since both cover the same distance,

$\frac{1}{2} a_{1} {(T - t)}^{2} = \frac{1}{2} a_{2} T^{2}$

So, $\sqrt{a_{1}} (T - t) = \sqrt{a_{2}} T$

Difference in their final speeds:

$v = a_{1} (T - t) - a_{2} T$

Using the above relation,

$v = \sqrt{a_{1} a_{2}} t$

Solution

Q.
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 of car B. Both the cars start from rest and travel with constant acceleration $a_{1}$ and $a_{2}$ respectively. Then $' v'$ is equal to:

1 $\frac{2 a_{1} a_{2}}{a_{1} + a_{2}} t$

2 $\sqrt{2 a_{1} a_{2}} t$

3 $\sqrt{a_{1} a_{2}} t$

4 $\frac{a_{1} + a_{2}}{2} t$

Want Us to Email you About Special Offers & Updates?

Solution

Q. 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 of car B. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then 'v' is equal to:

1 2a1a2a1+a2t

2 2a1a2t

3 a1a2t

4 a1+a22t