Q124 / 160 · TS EAMCET 2023 (13 May Shift 2) · Physics
A car is travelling with linear velocity ' $\mathrm{V}$ ' on a circular road of radius ' $r$ '. If its velocity is increasing at a rate of ' $a$ ' $\mathrm{ms}^{-2}$, then the resultant acceleration will be
- A$\sqrt{\left(\frac{\mathrm{v}^2}{\mathrm{r}^2}-\mathrm{a}^2\right)}$
- Correct answerB$\sqrt{\left(\frac{\mathrm{v}^4}{\mathrm{r}^2}+\mathrm{a}^2\right)}$
- C$\sqrt{\left(\frac{\mathrm{v}^4}{\mathrm{r}^2}-\mathrm{a}^2\right)}$
- D$\sqrt{\left(\frac{\mathrm{V}^2}{\mathrm{r}^2}+\mathrm{a}^2\right)}$
Solution
Radial acceleration, $a_r=\frac{V^2}{r}$
Tangential acceleration, $a_t=a$
Resultant acceleration, $a_R=\sqrt{a_T^2+a_t^2}$
$$
\begin{aligned}
& =\sqrt{\left(\frac{v^2}{r}\right)^2+a^2} \\
& =\sqrt{\frac{v^4}{r^2}+a^2}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.