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A thief is running away on a straight road in the jeep moving with a speed of 9 m/s. A policeman on a motorcycle chases him at a speed of 10 m/s. If at any instant the separation between the jeep and the motorcycle is 100 m, then in what time does the policeman catch the thief?
A) 1s
B) 9s
C) 10s
D) 100s

Last updated date: 14th Apr 2024
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MVSAT 2024
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Hint: Given problem is an example of a relative motion in a straight line. Just think how will the thief observe the policeman? Will the policeman move at 10 m/s or at a lower speed w.r.t the thief? Make the thief as the frame of reference and try to solve the problem.

Formula Used:
If two things or persons are in a motion in the same direction with different-different speeds and speed of one (let’s say \[u\]) is more than the speed of another (let’s say \[v\]).
Then the relative speed of one with respect to another is \[\mathop V\nolimits_{rel} = (u - v)m/s\]

Complete step by step answer:
 Speed of policeman on motor cycle = 10 m/s
             Speed of thief in the jeep = 9 m/s
            The relative speed of policeman w.r.t thief is calculated i.e.
             Speed of policeman on motorcycle - Speed of thief in the jeep i.e.
              \[\mathop V\nolimits_{rel} = \](10-9) m/s =1m/s
Step 2: We know that, at some instant separation between policeman and thief i.e. \[S = \]100m
Then time taken by the policeman to catch the thief i.e.

\[t = \dfrac{S}{{\mathop V\nolimits_{rel} }}\]

\[t = \dfrac{{100}}{1} = 100s\]

The correct option is (D).

Note: This problem can also be solved from earth’s frame of reference too. Starting with the initial 100m gap between police and thief and considering police’s initial position as origin, write the equation of distance covered for both police and thief and equate them at time t(when police will catch the thief and hence their distance will be the same from origin).