A particle experiences a constant acceleration for $20 \,sec$ after starting from rest. If it travels a distance ${S_1}$ in the first $10\, sec$ and a distance ${S_2}$ in the next $10 \,sec$, then
${S_1} = {S_2}$
${S_1} = {S_2}/3$
${S_1} = {S_2}/2$
${S_1} = {S_2}/4$
A particle experiences a constant acceleration for $20 \,sec$ after starting from rest. If it travels a distance ${S_1}$ in the first $10\, sec$ and a distance ${S_2}$ in the next $10 \,sec$, then
As $S = ut + \frac{1}{2}a{t^2}$
$\therefore {S_1} = \frac{1}{2}a{(10)^2} = 50a$ .....(i)
$As\;\;v = u + at$
$\therefore $ velocity acquired by particle in $10 \,sec$ $v = a \times 10$
For next $10\, sec$ ,
${S_2} = (10a) \times 10 + \frac{1}{2}(a) \times {(10)^2}$
${S_2} = $ $150a$ .....(ii)
From (i) and (ii)
${S_1} = {S_2}/3$
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