Question

A body travels for 15 sec starting from rest with constant acceleration. If it travels distances $S_{1},\mspace{6mu} S_{2}$ and $S_{3}$ in the first five seconds, second five seconds and next five seconds respectively the relation between $S_{1},\mspace{6mu} S_{2}$ 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)

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

Answer 2

Since the body starts from rest. Therefore $u = 0$.

$S_{1} = \frac{1}{2}a(5)^{2} = \frac{25a}{2}$

$\mathbf{S}_{\mathbf{1}}\mathbf{+}\mathbf{S}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{a(10}\mathbf{)}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{100a}}{\mathbf{2}}$ ⇒ $S_{2} = \frac{100a}{2} - S_{1}$=$75\frac{a}{2}$

$S_{1} + S_{2} + S_{3} = \frac{1}{2}a(15)^{2} = \frac{225a}{2}$

⇒ $S_{3} = \frac{225a}{2} - S_{2} - S_{1}$= $\frac{125a}{2}$

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