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Standard IX

Physics

Velocity

Two cars A ...

Question

Two cars $A$ and $B$ are at rest at same point initially. If $A$ starts with a uniform velocity of $40 m / s e c$ and $B$ starts in the same direction with a constant acceleration of $4 m / s^{2}$ , then $B$ will catch $A$ after how much time?

10 s e c

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20 s e c

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30 s e c

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35 s e c

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Solution

The correct option is B $20 s e c$
Let distance covered by A be $S_{1}$
Let distance covered by B be $S_{2}$
B will catch A when $S_{1} = S_{2}$
$S = u t + 1 / 2 \times a \times t^{2}$

Therefore from given data,
$40 t + (1 / 2 \times 0 \times t^{2}) = 0 t + 1 / 2 \times 4 \times t^{2}$
$40 t = 2 t^{2}$
$20 t = t^{2}$
$t = 20 s$

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Two cars A and B are at rest at same point initially. If A starts with a uniform velocity of 40 m/sec and B starts in the same direction with a constant acceleration of 4 m/s2, then B will catch A after how much time?

The correct option is B 20 secLet distance covered by A be S1Let distance covered by B be S2B will catch A when S1=S2S=ut+1/2×a×t2Therefore from given data,40t+(1/2×0×t2)=0t+1/2×4×t240t=2t220t=t2t=20s

Two cars $A$ and $B$ are at rest at same point initially. If $A$ starts with a uniform velocity of $40 m / s e c$ and $B$ starts in the same direction with a constant acceleration of $4 m / s^{2}$ , then $B$ will catch $A$ after how much time?