Question
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 :-
$\mathrm{S} 1=\mathrm{ut}+1 / 2 \times \mathrm{at}^{\wedge} 2$
$\mathrm{S} 1=0 \times 10+1 / 2 \times \mathrm{a} \times 100$
$\mathrm{So}, \mathrm{S} 1=50 \mathrm{a}$
Now, $\mathrm{S} 1+\mathrm{S} 2=0 \times 20+1 / 2 \times \mathrm{a} \times 20^{\wedge} 2$
$\mathrm{S} 1+\mathrm{S} 2=200 \mathrm{a}$
As, $\mathrm{S} 1=50 \mathrm{a}$
$50 a+52=200 a$
$\mathrm{S} 2=150 \mathrm{a}$
Hence $\mathrm{S} 1 / \mathrm{S} 2=50 \mathrm{a} / 150 \mathrm{a}$
$\mathrm{S} 2=3 \mathrm{S} 1$
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$Reason$ : A body is numerically at rest when it reverses its direction.