A bus starts from rest and moves with a uniform acceleration of 1ms−2. A boy 10m behind the bus at the start runs at a constant speed and catches the bus in 10s. Speed of the boy is :
10ms−1
1ms−1
6ms−1
4ms−1
A
1ms−1
B
4ms−1
C
6ms−1
D
10ms−1
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Solution
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Solution 2
Distance travelled
by bus in 10s⇒12×1×102=50m D = Distance boy has to cover =10+50=60m Speed
of boy =DT=6010=6m/s
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