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An armored car 2m long and 3m wide is moving at 10ms^–1 when a bullet hits it in a direction making an angle $θ = \tan^{- 1} (\frac{3}{4})$ with the length of the car as seen by a stationary observer. The bullet enters at one edge of the car at the corner and passes out at the diagonally opposite corner. Neglecting any interaction between the car and the bullet and effect of gravity, the time for the bullet to cross the car is

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0.15 sec

0.5 sec

0.1 sec

0.2 sec

answer is A.

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Detailed Solution

Given $θ = \tan^{- 1} (3 / 4)$

$θ = 37^{0}$

Components of velocity of bullet will be $v \sin θ$ in the direction of the car and

$v \cos θ$ perpendicular to the direction of the car.

Therefore,

(V_b cos37^o - V_c)t = 2

(V_b sin37^o)t = 3

Solving will give us t = 0.2 secs

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