Question

A particle starts its motion from rest under the action of a constant force. If the distance covered in first $10seconds$ is $S_1$ and that covered in the first $20seconds$ is $S_2$ then

• A. $S_2=2S_1$
• B. $S_2=3S_1$
• C. $S_2=4S_1$
• D. $S_2=S_1$

Answer 1

The correct option is C

$S_2=4S_1$

Step 1. Given data:

$S_1=$Distance covered in first $10$ seconds

$S_2=$Distance covered in first $20$ seconds

$t_1=10$, Time taken during $S_1$ distance travelled

$t_2=20$, Time taken during $S_2$ distance travelled

Step 2. Calculating $S_1$ and $S_2$ using equation of motion:

From the second equation of motion,

As, $S=ut+\frac{1}{2}a{t}^2$ $\left[u=0\right]$

$$\begin{gathered} S_1=\frac{1}{2}a{\left(t_1\right)}^2 \\ \Rightarrow S_1=\frac{1}{2}a{\left(10\right)}^2 \end{gathered}$$

$\therefore S_1=50a$ _____________ $\left(1\right)$

And

$$\begin{gathered} S_2=\frac{1}{2}a{\left(t_2\right)}^2 \\ \Rightarrow S_2=\frac{1}{2}a{\left(20\right)}^2 \end{gathered}$$

$\therefore S_2=200a$ _____________ $\left(2\right)$

Dividing equation $\left(1\right)$ by $\left(2\right)$

$\begin{array}{rcl}\frac{S_1}{S_2}& =& \frac{50a}{200a}\\ & \Rightarrow & \frac{S_1}{S_2}=\frac{1}{4}\end{array}$

$\therefore S_2=4S_1$

Hence, the correct option is (C).