| A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out a length\[s={{t}^{3}}+5,\]where s is in metres and t is in seconds. The radius of the path is 20 m. The acceleration of P when\[t=2s\] is nearly. [AIEEE 2010] |
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A) \[14\text{ }m/{{s}^{2}}\]
B) \[13\text{ }m/{{s}^{2}}\]
C) \[12\text{ }m/{{s}^{2}}\]
D) \[7.2\text{ }m/{{s}^{2}}\]
Correct Answer: C
Solution :
| [a] \[s={{t}^{3}}+5\] |
| \[v=\frac{ds}{dt}=3{{t}^{2}}\] |
| \[\frac{dv}{dt}=6t\] |
| \[a=\sqrt{(a_{r}^{2}+a_{t}^{2})}\]\[=\sqrt{{{\left( \frac{{{v}^{2}}}{R} \right)}^{2}}+{{\left( \frac{dv}{dt} \right)}^{2}}}=14\,m/{{s}^{2}}\] |
| At \[t=2s\] |
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