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JEE MAIN 2019
09-01-19 S2
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:
Select the correct option:
A
$\frac{a_1+a_2}{2} t$
B
$\frac{2 a_1 a_2}{a_1+a_2} t$
C
$\sqrt{2 a_1 a_2 t}$
D
$\sqrt{a_1 a_2} t$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$ \begin{aligned} & t_A=t_B-t \\ & v_A=a_1\left(t_B-t\right)=a_2 t_B+v \\ & S=\frac{1}{2} a_1\left(t_B-t\right)^2=\frac{1}{2} a_2 t_B^2 \\ & \Rightarrow t_B\left[1-\sqrt{\frac{a_2}{a_1}}\right]=t \end{aligned} $$ Solving (i) and (ii) $\mathrm{v}=\sqrt{\mathrm{a}_1 \mathrm{a}_2 \mathrm{t}}$
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