Question

A body travels for $15\sec $ 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. ${s_1} = {s_2} = {s_3}$
B. $5{s_1} = 3{s_2} = {s_3}$
C. \[{s_1} = \dfrac{1}{3}{s_2} = \dfrac{1}{5}{s_3}\]
D. ${s_1} = \dfrac{1}{5}{s_2} = \dfrac{1}{3}{s_3}$

Answer 1

Hint:Here we will take three intervals of time and take the initial velocity of the body to be zero and apply an equation of motion to calculate the distances. After finding the distances of each interval we can find the relation between ${s_1},{s_2}$ and ${s_3}$ .

Formula used:
$S = ut + \dfrac{1}{2}a{t^2}$
Where, $S$ is the distance, $u$ is the initial speed, $t$ is the time and $a$ is the acceleration.

Complete step by step answer:
Let $v$ denote speed and $a$ denote acceleration.According to the question,
For interval-1 from $t = 0$ to $t = 5$
At, $t = 0$ body is at rest so $v = 0$
Now, putting the values in formula,
$\because S = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow {S_1} = u(0) + \dfrac{1}{2}a({5^2}) \\
\Rightarrow {S_1} = 12.5a \\ $
Similarly, for interval-2 from $t = 5$ to $t = 10$
At, $t = 5$ , $v = 5$
Now, putting the values in formula,
$\because S = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow {S_2} = 5 \times 5 + \dfrac{1}{2}a({5^2}) \\
\Rightarrow {S_2} = 37.5a \\ $
Similarly, for interval- from $t = 10$ to $t = 15$
At, $t = 10$ , $v = 10$
Now, putting the values in formula,
$\because S = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow {S_3} = 10 \times 5 + \dfrac{1}{2}a({5^2}) \\
\Rightarrow {S_3} = 62.5a \\ $
Now, we have ${S_1} = 12.5a$ , ${S_2} = 37.5a$ and ${S_3} = 62.5a$
So, the relation between ${s_1},{s_2}$ and ${s_3}$ is,
${s_1}:{s_2}:{s_3}$
$\Rightarrow 12.5a:37.5a:62.5a \\
\Rightarrow 1:3:5 \\ $
$\Rightarrow {s_1}:{s_2}:{s_3} = 1:3:5 \\
\therefore \dfrac{{{s_1}}}{1} = \dfrac{{{s_2}}}{3} = \dfrac{{{s_3}}}{5} \\ $
So, the relation between ${s_1},{s_2}$ and ${s_3}$ is, \[{s_1} = \dfrac{1}{3}{s_2} = \dfrac{1}{5}{s_3}\] .

Hence the correct option is C.

Note:In the alternative way, students can solve this question by calculating the total distance travelled by the particle starting from the rest in 15 seconds. Then calculate the distance travelled in the first 5 seconds and then 10 seconds and subtract it from the total distance travelled. Note that, in our solution, the initial velocity of the particle for the second interval is not zero.