A bus starts from rest and moves with a uniform acceleration of $1 m s^{- 2}$ . A boy $10 m$ behind the bus at the start runs at a constant speed and catches the bus in $10 s$ . Speed of the boy is :

10 m s^{- 1}

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1 m s^{- 1}

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6 m s^{- 1}

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4 m s^{- 1}

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Solution

The correct option is C $6 m s^{- 1}$

Solution 2
Distance travelled by bus in $10 s \Rightarrow \frac{1}{2} \times 1 \times 10^{2} = 50 m$
D $=$ Distance boy has to cover $= 10 + 50 = 60 m$
Speed of boy $= \frac{D}{T} = \frac{60}{10} = 6 m / s$

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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 :

The correct option is C 6ms−1 Solution 2Distance travelled by bus in 10s⇒12×1×102=50mD = Distance boy has to cover =10+50=60mSpeed of boy =DT=6010=6 m/s

A bus starts from rest and moves with a uniform acceleration of $1 m s^{- 2}$ . A boy $10 m$ behind the bus at the start runs at a constant speed and catches the bus in $10 s$ . Speed of the boy is :

The correct option is C $6 m s^{- 1}$

Solution 2
Distance travelled by bus in $10 s \Rightarrow \frac{1}{2} \times 1 \times 10^{2} = 50 m$
D $=$ Distance boy has to cover $= 10 + 50 = 60 m$
Speed of boy $= \frac{D}{T} = \frac{60}{10} = 6 m / s$