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(C) For a body starting from rest $(u = 0)$ with constant acceleration $(a)$,the distance traveled in time $t$ is given by $S = \frac{1}{2}at^2$.Let t

Vedclass · Accepted Answer

${S_1} = \frac{1}{3}{S_2} = \frac{1}{5}{S_3}$

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(C) For a body starting from rest $(u = 0)$ with constant acceleration $(a)$,the distance traveled in time $t$ is given by $S = \frac{1}{2}at^2$.Let the time interval be $T = 5\, s$.Distance in the first $5\, s$ $(t=T)$: ${S_1} = \frac{1}{2}aT^2$.Distance in the first $10\, s$ $(t=2T)$: ${S_1} + {S_2} = \frac{1}{2}a(2T)^2 = 4 	imes (\frac{1}{2}aT^2) = 4{S_1}$. Thus,${S_2} = 3{S_1}$.Distance in the first $15\, s$ $(t=3T)$: ${S_1} + {S_2} + {S_3} = \frac{1}{2}a(3T)^2 = 9 	imes (\frac{1}{2}aT^2) = 9{S_1}$. Thus,${S_3} = 9{S_1} - ({S_1} + {S_2}) = 9{S_1} - 4{S_1} = 5{S_1}$.Therefore,the ratio is ${S_1}:{S_2}:{S_3} = 1:3:5$.This implies ${S_1} = \frac{1}{3}{S_2} = \frac{1}{5}{S_3}$.

$A$ body travels for $15\, s$ starting from rest with constant acceleration. If it travels distances ${S_1}$,${S_2}$,and ${S_3}$ in the first $5\, s$,the second $5\, s$,and the next $5\, s$ respectively,the relation between ${S_1}$,${S_2}$,and ${S_3}$ is:

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