(D) The speed of the thief's jeep is $v_t = 9 \, m/s$.The speed of the policeman's motorcycle is $v_p = 10 \, m/s$.The relative speed of the policeman

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(D) The speed of the thief's jeep is $v_t = 9 \, m/s$.The speed of the policeman's motorcycle is $v_p = 10 \, m/s$.The relative speed of the policeman with respect to the thief is $v_{rel} = v_p - v_t = 10 - 9 = 1 \, m/s$.The initial distance (separation) between them is $d = 100 \, m$.To catch the thief,the policeman must cover this relative distance at the relative speed.Time taken $t = \frac{d}{v_{rel}} = \frac{100 \, m}{1 \, m/s} = 100 \, s$.Therefore,it will take $100 \, seconds$ for the policeman to catch the thief.

Class 11 Physics Motion in Straight Line Relative Motion in One Dimension

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$A$ thief is running away on a straight road in a jeep moving with a speed of $9 \, m/s$. $A$ policeman chases him on a motorcycle moving at a speed of $10 \, m/s$. If the instantaneous separation between the jeep and the motorcycle is $100 \, m$,how long will it take for the policeman to catch the thief?

A
$1$
B
$19$
C
$90$
D
$100$

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Motion in Straight Line

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Relative Motion in One Dimension

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