A body starting from rest is moving under a constant acceleration uo to 20 sec. if it moves $S_{1}$ distance in first 10 sec., and $S_{2}$ distance in next 10 sec. then $S_{2}$ will be equal to :

S_{1}

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2 S_{1}

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3 S_{1}

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4 S_{1}

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Solution

The correct option is C $3 S_{1}$
$a = a$
$t = 10 s$
$S_{10} = S_{1} = \frac{1}{2} a t^{2} = \frac{1}{2} a (10)^{2}$
$= 50 a$
$v_{1} = a t = 10 a$
$t = 20 s$
$S_{20} (S_{2} + S_{1}) = \frac{1}{2} a t^{2} = \frac{1}{2} a (20)^{2} = 200 a$
$S_{2} = 200 a - S_{1} = 150 a$
$\frac{S_{2}}{S_{1}} = 3$ .

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A body starting from rest is moving under a constant acceleration uo to 20 sec. if it moves S1 distance in first 10 sec., and S2 distance in next 10 sec. then S2 will be equal to :

The correct option is C 3S1a=at=10sS10=S1=12at2=12a(10)2=50av1=at=10at=20sS20(S2+S1)=12at2=12a(20)2=200aS2=200a−S1=150aS2S1=3.

A body starting from rest is moving under a constant acceleration uo to 20 sec. if it moves $S_{1}$ distance in first 10 sec., and $S_{2}$ distance in next 10 sec. then $S_{2}$ will be equal to :

The correct option is C $3 S_{1}$
$a = a$
$t = 10 s$
$S_{10} = S_{1} = \frac{1}{2} a t^{2} = \frac{1}{2} a (10)^{2}$
$= 50 a$
$v_{1} = a t = 10 a$
$t = 20 s$
$S_{20} (S_{2} + S_{1}) = \frac{1}{2} a t^{2} = \frac{1}{2} a (20)^{2} = 200 a$
$S_{2} = 200 a - S_{1} = 150 a$
$\frac{S_{2}}{S_{1}} = 3$ .