Question

A body starting from rest is travelling on a straight road with constant non-zero acceleration. If the speeds after covering distances $S_1$ and $S_2$ be $V_1$ and $V_2$ respectively. If $V_2/V_1=2$, then $S_2/S_1=N$. Find $N$.

Answer 1

Solution

(i) While covering $S_1$ distance

$u=0m/s$ and $V=V_1$, so,

Applying third $e{q}^n$ of motion,

$(V_1{)}^2-(0{)}^2=2\left(S_1\right)\left(a\right)$

$\Rightarrow (V_1{)}^2=2S_{1}a\Rightarrow V_1=\sqrt{2S_{1}a}$

(ii) While covering $S_2$ distance,

$u=V_{1}m/s$ and $V=V_{2}m/s$. so,

Applying third $e{q}^n$ of motion,

$(V_2{)}^2-(V_1{)}^2=2\left(S_2\right)\left(a\right)$

$\Rightarrow (V_2{)}^2=2S_{2}a+V_1^2$

$\Rightarrow V_2^2=2(S_1+S_2)a$

$\Rightarrow V_1=\sqrt{2(S_1+S_2)}a$

As $\frac{V_2}{V_1}=2$ Let $V_1=K,$ then $V_2=2K$

so, $V_1=\sqrt{2S_{1}a}\Rightarrow S_1=\frac{V_1^2}{2a}\Rightarrow S_1=\frac{K^2}{2a}$

Also, $S_1+S_2=\frac{V_2^2}{2a}\Rightarrow \frac{K^2}{2a}+S_2=\frac{V_2^2}{2a}$

$\Rightarrow S_2=\frac{4K^2}{2a}-\frac{K^2}{2a}=\frac{3K^2}{2a}$

So, $\frac{S_2}{S_1}=3\Rightarrow N$ is 3.