A particle experiences constant acceleration for $20$ seconds after starting from rest. If it travels a distance $s_{1}$ in the first $10$ seconds and distance $s_{2}$ in the next $10$ second then:

s_{2} = s_{1}

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s_{2} = 2 s_{1}

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s_{2} = 3 s_{1}

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s_{2} = 4 s_{1}

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Solution

The correct option is C $s_{2} = 3 s_{1}$
Distance traveled in first $10 s e c o n d s$

$S_{1} = u t + \frac{1}{2} a t^{2} = 0 t + \frac{1}{2} a 10^{2} = 50 a$

Total distance traveled in 20 seconds =

$S = u t + \frac{1}{2} a t^{2} = 0 t + \frac{1}{2} a 2062 = 200 a$

Distance traveled from $10 s$ to $20 s$ $= S_{2} = 200 a - 50 a = 150 a$
Hence answer is C.

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