MHT CET · Physics · Motion In Two Dimensions
A body is moving along a circular track of radius \(100 \mathrm{~m}\) with velocity \(20 \mathrm{~m} / \mathrm{s}\). Its
tangential acceleration is \(3 \mathrm{~m} / \mathrm{s}^{2}\), then its resultant acceleration will be
- A \(3 \mathrm{~m} / \mathrm{s}^{2}\)
- B \(5 \mathrm{~m} / \mathrm{s}^{2}\)
- C \(4 \mathrm{~m} / \mathrm{s}^{2}\)
- D \(2 \mathrm{~m} / \mathrm{s}^{2}\)
Answer & Solution
Correct Answer
(B) \(5 \mathrm{~m} / \mathrm{s}^{2}\)
Step-by-step Solution
Detailed explanation
(A)
Tangential acceleration \(\mathrm{a}_{\mathrm{t}}=3 \mathrm{~m} / \mathrm{s}^{2}\)
radial acceleration \(\mathrm{a}_{\mathrm{r}}=\frac{\mathrm{V}^{2}}{\mathrm{r}}=\frac{(20)^{2}}{100}=4 \mathrm{~m} / \mathrm{s}^{2}\)
Total acceleration \(=\sqrt{a_{r}^{2}+a_{C}^{2}}=\sqrt{(4)^{2}+(3)^{2}}=5 \mathrm{~m} / \mathrm{s}^{2}\)
Tangential acceleration \(\mathrm{a}_{\mathrm{t}}=3 \mathrm{~m} / \mathrm{s}^{2}\)
radial acceleration \(\mathrm{a}_{\mathrm{r}}=\frac{\mathrm{V}^{2}}{\mathrm{r}}=\frac{(20)^{2}}{100}=4 \mathrm{~m} / \mathrm{s}^{2}\)
Total acceleration \(=\sqrt{a_{r}^{2}+a_{C}^{2}}=\sqrt{(4)^{2}+(3)^{2}}=5 \mathrm{~m} / \mathrm{s}^{2}\)
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