A thief runs at a speed of 5 , m/s. A policem | vedclass

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Question

(B) Let the time taken by the policeman to catch the thief be $t$ seconds.Distance covered by the thief in $(t + 2)$ seconds $= 5 	imes (t + 2) \, m$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let the time taken by the policeman to catch the thief be $t$ seconds.Distance covered by the thief in $(t + 2)$ seconds $= 5 	imes (t + 2) \, m$.Distance covered by the policeman in $t$ seconds $= 10 	imes t \, m$.Since the policeman catches the thief,the distances covered by both must be equal:$10t = 5(t + 2)$$10t = 5t + 10$$5t = 10$$t = 2 \, seconds$.Alternatively,using relative speed:The thief has a head start of $2 \, seconds 	imes 5 \, m/s = 10 \, m$.Relative speed $= 10 - 5 = 5 \, m/s$.Time taken to cover the gap $= \frac{10 \, m}{5 \, m/s} = 2 \, seconds$.

$A$ thief runs at a speed of $5 \, m/s$. $A$ policeman runs behind him $2 \, seconds$ after the thief started at a speed of $10 \, m/s$. After how much time will the policeman catch the thief? (in $seconds$)

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