TS EAMCET · Physics · Motion In Two Dimensions
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 \(\sqrt{\left(\frac{\mathrm{v}^2}{\mathrm{r}^2}-\mathrm{a}^2\right)}\)
- 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 \(\sqrt{\left(\frac{\mathrm{V}^2}{\mathrm{r}^2}+\mathrm{a}^2\right)}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\left(\frac{\mathrm{v}^4}{\mathrm{r}^2}+\mathrm{a}^2\right)}\)
Step-by-step Solution
Detailed explanation
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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