Question

A body travels for $15sec$ 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, the relation between $s_1,s_2$ and $s_3$ is

• A. $s_1=\frac{1}{3}{s}_2=\frac{1}{5}{s}_3$
• B. $s_1=\frac{1}{5}{s}_2=\frac{1}{3}{s}_3$
• C. $2s_1=\frac{1}{3}{s}_2=\frac{1}{5}{s}_3$
• D. $s_1=\frac{1}{6}{s}_2=\frac{1}{5}{s}_3$

Answer 1

The correct option is A $s_1=\frac{1}{3}{s}_2=\frac{1}{5}{s}_3$

Since the body starts from rest. Therefore, $u=0$
$\Rightarrow s_1=\frac{1}{2}a(5{)}^2=\frac{25a}{2}$
and,
$s_1+s_2=\frac{1}{2}a(10{)}^2=\frac{100a}{2}\Rightarrow s_2=\frac{100a}{2}-s_1=75\frac{a}{2}$

Also,
$s_1+s_2+s_3=\frac{1}{2}a(15{)}^2=\frac{225a}{2}\Rightarrow s_3=\frac{225a}{2}-s_2-s_1=\frac{125a}{2}$

$\therefore$ taking the ratio for all three distances, we get $s_1=\frac{1}{3}{s}_2=\frac{1}{5}{s}_3$