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Question

As acceleration is constant for 20 sec after starting from rest so

$\mathrm{S} 1=\mathrm{ut}+1 / 2 	imes \mathrm{at}^{\wedge} 2$

$\mathrm{S} 1=

Vedclass · Accepted Answer

$s_2 = 3s_1$

Vedclass · Answer

As acceleration is constant for 20 sec after starting from rest so

$\mathrm{S} 1=\mathrm{ut}+1 / 2 	imes \mathrm{at}^{\wedge} 2$

$\mathrm{S} 1=0 	imes 10+1 / 2 	imes \mathrm{a} 	imes 100$

$\mathrm{So}, \mathrm{S} 1=50 \mathrm{a}$

Now, $\mathrm{S} 1+\mathrm{S} 2=0 	imes 20+1 / 2 	imes \mathrm{a} 	imes 20^{\wedge} 2$

$\mathrm{S} 1+\mathrm{S} 2=200 \mathrm{a}$

As, $\mathrm{S} 1=50 \mathrm{a}$

$50 a+52=200 a$

$\mathrm{S} 2=150 \mathrm{a}$

Hence $\mathrm{S} 1 / \mathrm{S} 2=50 \mathrm{a} / 150 \mathrm{a}$

$\mathrm{S} 2=3 \mathrm{S} 1$

A particle experiences a 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 $10\, seconds$, then :-

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