In a car race, car A taken 2 s less then car B and passes the finishing point with a velocity v more than the velocity with which car B passes the point. Both the cars start from rest and travel with constant accelerations of $a_{A} = 3 {ms}^{- 2} and a_{B} = 2 {ms}^{- 2}$ . The value of v is

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$2 .45 {ms}^{- 1}$

$4 .90 {ms}^{- 1}$

$7 .35 {ms}^{- 1}$

$9 .80 {ms}^{- 1}$

answer is B.

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Detailed Solution

Given data $t_{B} = t_{A} + 2$

$\begin{array}{l} t_{B} - t_{A} = 2 \\ v_{A} = v_{B} + v \\ v_{A} - v_{B} = v \\ \frac{v_{A} - v_{B}}{t_{B} - t_{A}} = \frac{v}{2} \\ \frac{v}{2} = \frac{\sqrt{2 a_{A} s} - \sqrt{2 a_{B} s}}{\sqrt{\frac{2 s}{a_{B}}} - \sqrt{\frac{2 s}{a_{A}}}} \\ \frac{v}{2} = \frac{\sqrt{2 \times 3} - \sqrt{2 \times 2}}{\sqrt{\frac{2}{2}} - \sqrt{\frac{2}{3}}} \end{array}$

$\begin{array}{l} \frac{v}{2} = \frac{\sqrt{6} - 2}{1 - \frac{\sqrt{2}}{\sqrt{3}}} \\ \frac{v}{2} = \frac{(\sqrt{6} - 2) \sqrt{3}}{\sqrt{3} - \sqrt{2}} \\ \frac{v}{2} = (\sqrt{6} - 2) \sqrt{3} (\sqrt{3} - \sqrt{2}) \\ = (2.44 - 2) 1.732 (1.732 + 1.414) \\ = 0.44 \times 1.732 \times 3.146 \\ \frac{v}{2} = 2.3975 \\ v = 2 \times 2.3975 \\ v = 4.79 {ms}^{- 1} \end{array}$