Question

A body travels for 15 seconds starting from rest with constant acceleration. If it travels distances $S_1,S_2$ and $S_3$ in the first five seconds, second five seconds and next five seconds respectively, then the relation between $S_1,S_2$ and $S_3$ is

• A. $S_1=S_2=S_3$
• B. $5S{1}_{=}3S_2=S_3$
• C. $S_1=\frac{1}{3}{S}_2=\frac{1}{5}{S}_3$
• D. $S_1=\frac{1}{5}{S}_2=\frac{1}{3}{S}_3$

Answer 1

Answer: (C)

Let a be uniform acceleration of the body.
As $S=ut+\frac{1}{2}a{t}^2=\frac{1}{2}a{t}^2\left(\because u=0\right)$
Then, $S_1=\frac{1}{2}a{\left(5\right)}^2...\left(i\right)$
$S_1+S_2=\frac{1}{2}a{\left(10\right)}^2...\left(ii\right)$
$S_1+S_2+S_3=\frac{1}{2}a{\left(15\right)}^2...\left(iii\right)$
Subtract $\left(i\right)$ from $\left(ii\right)$, we get
$\left(S_1+S_2\right)-S_1=\frac{1}{2}a{\left(10\right)}^2-\frac{1}{2}a{\left(5\right)}^2$
$S_2=\frac{75}{2}a=3S_1$ (Using $\left(i\right)$)
Subtract $\left(ii\right)$ from $\left(iii\right)$, we get
$\left(S_1+S_2+S_3\right)-\left(S_1+S_2\right)=\frac{1}{2}a{\left(15\right)}^2-\frac{1}{2}a{\left(10\right)}^2$
$S_3=\frac{125}{2}a=5S_1$ (Using $\left(i\right)$)
Thus, $S_1=\frac{1}{3}{S}_2=\frac{1}{5}{S}_3$