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. $5S1_ = 3S_2 = S_3$
• 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)

$S_{1} = \frac{1}{3} S_{2} = \frac{1}{5}S_{3}$

Answer 2

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