Question

At some instant $\overrightarrow{v}=4\hat{i}-3\hat{j}\text{m/s}$ and $\overrightarrow{a}=2\hat{i}+\hat{j}{\text{m/s}}^2$. Find the radius of curvature at that instant.

• A. $12.5\text{m}$
• B. $25\text{m}$
• C. $15\text{m}$
• D. $20\text{m}$

Answer 1

The correct option is A $12.5\text{m}$

Given, $\overrightarrow{v}=4\hat{i}-3\hat{j}$ and $\overrightarrow{a}=2\hat{i}+\hat{j}$
We have, $|\overrightarrow{v}|=\sqrt{4^2+(-3{)}^2}=5\text{m/s}$
and, $|\overrightarrow{a}|=\sqrt{2^2+1^2}=\sqrt{5}{\text{m/s}}^2$
Also, we know that the component of acceleration parallel to the velocity is given as
$a_{\parallel }=\frac{\overrightarrow{a}\cdot \overrightarrow{v}}{|\overrightarrow{v}|}=\frac{(2\hat{i}+\hat{j})\cdot (4\hat{i}-3\hat{j})}{5}=\frac{8-3}{5}=1{\text{m/s}}^2$
Let us say that the component of acceleration perpendicular to the velocity be $a_t$
So, we have
$|\overrightarrow{a}{|}^2=a_t^2+a_{\parallel }^2$
$\Rightarrow (\sqrt{5}{)}^2=a_t^2+1^2\Rightarrow a_t=2{\text{m/s}}^2$
Thus, radius of curvature at that instant is given by
$R_c=\frac{|\overrightarrow{v}{|}^2}{a_t}=\frac{5^2}{2}=12.5\text{m}$