A body starts from rest and is uniformly accelerated for $30 \mathrm{s}$. The distance travelled in the first $10 \mathrm{s}$ is $S_1$, next $10 \mathrm{s}$ is $S_2$ and the last 10 s is $S_3$. Then $S_1 : S_{2 } : S_3$ is the same as

• A

  $1 : 2 : 4$

   
• B

  $1 : 2 : 5$

   
• Correct answer
  C

  $1 : 3 : 5$

   
• D

  $1 : 3 : 9$

   

Solution

a = constant

$S_1=v_{0}t+\frac{1}{2}a{t}^2$

$v_1=v_0+at$

${\mathit{S}}_2={\mathit{\text{v}}}_1\mathit{\text{t}}+\frac{1}{2}{\mathit{\text{at}}}^2$ $\Rightarrow v_{0}t+\frac{3}{2}a{t}^2$

${\mathit{\text{v}}}_2={\mathit{\text{v}}}_1+\mathit{\text{at}}$

$={\mathit{\text{v}}}_0+2\mathit{\text{at}}$

${\mathit{\text{S}}}_3={\mathit{\text{v}}}_2\mathit{\text{t}}+\frac{1}{2}{\mathit{\text{at}}}^2$

$={\mathit{\text{v}}}_0\mathit{\text{t}}+\frac{5}{2}{\mathit{\text{at}}}^2$

${\mathit{\text{v}}}_0=0$

${\mathit{\text{S}}}_1=\frac{1}{2}{\mathit{\text{at}}}^2$;       ${\mathit{\text{S}}}_2=\frac{3}{2}{\mathit{\text{at}}}^2$;      ${\mathit{\text{S}}}_3=\frac{5}{2}{\mathit{\text{at}}}^2$

${\mathit{\text{S}}}_1\text{ : }{\mathit{\text{S}}}_2\text{ : }{\mathit{\text{S}}}_3=\text{1}\text{ : }3\text{ : }5$

Hence, the correct answer is option: 'C'.