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Question: Answered & Verified by Expert
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
PhysicsMotion In Two DimensionsTS EAMCETTS EAMCET 2023 (13 May Shift 2)
Options:
  • A $\sqrt{\left(\frac{\mathrm{v}^2}{\mathrm{r}^2}-\mathrm{a}^2\right)}$
  • B $\sqrt{\left(\frac{\mathrm{v}^4}{\mathrm{r}^2}+\mathrm{a}^2\right)}$
  • 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:
2765 Upvotes Verified Answer
The correct answer is: $\sqrt{\left(\frac{\mathrm{v}^4}{\mathrm{r}^2}+\mathrm{a}^2\right)}$
Radial acceleration, $a_r=\frac{V^2}{r}$
Tangential acceleration, $a_t=a$
Resultant acceleration, $a_R=\sqrt{a_T^2+a_t^2}$
$$
\begin{aligned}
& =\sqrt{\left(\frac{v^2}{r}\right)^2+a^2} \\
& =\sqrt{\frac{v^4}{r^2}+a^2}
\end{aligned}
$$

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