Question

In a car race on a straight road, car A takes a time $t$ less than car B at the finish and passes the finishing point with a speed $v$ more than that of car B. Both the cars start from rest and travel with constant accelerations $a_1$ and $a_2$ respectively. Then $v$ is equal to

• A. $\dfrac{2a_1a_2}{a_1+a_2}$
• B. $\sqrt{2a_1a_2}\,t$
• C. $\dfrac{a_1+a_2}{2}\,t$
• D. $\sqrt{a_1a_2}\,t$

Hint:

When two bodies start from rest and cover the same distance with uniform acceleration, relate their times using $s=\tfrac{1}{2}at^2$ before comparing velocities.

Answer 1

The Correct Option is D

Step 1: Let the common distance of the race be $s$. Since both cars start from rest with constant acceleration: \[ s=\frac{1}{2}a_1 t_1^2=\frac{1}{2}a_2 t_2^2 \] where $t_1$ and $t_2$ are the times taken by cars A and B respectively.
Step 2: Given that car A finishes $t$ seconds earlier: \[ t_2 - t_1 = t \]
Step 3: From the distance equation: \[ a_1 t_1^2 = a_2 t_2^2 \Rightarrow \frac{t_2}{t_1}=\sqrt{\frac{a_1}{a_2}} \]
Step 4: Write $t_2 = t_1 + t$ and substitute: \[ \frac{t_1+t}{t_1}=\sqrt{\frac{a_1}{a_2}} \]
Step 5: Solving gives: \[ t_1=\frac{t}{\sqrt{\frac{a_1}{a_2}}-1} \]
Step 6: Final speeds at the finish line: \[ v_A = a_1 t_1,\quad v_B = a_2 t_2 \]
Step 7: Given $v = v_A - v_B$: \[ v = a_1 t_1 - a_2(t_1+t) \] Substituting $t_1$ and simplifying gives: \[ v=\sqrt{a_1a_2}\,t \]