A) \[\frac{{{\pi }^{2}}}{18}\]
B) \[\frac{{{\pi }^{2}}}{7}\]
C) \[\frac{-{{\pi }^{2}}}{9}\]
D) \[\frac{-{{\pi }^{2}}}{18}\]
Correct Answer: D
Solution :
Given, \[s=\frac{2}{9}\sin \,\frac{\pi t}{2}+{{s}_{0}}\] \[\Rightarrow \] \[\frac{ds}{dt}=\frac{2}{9}\cos \frac{\pi t}{2}\times \frac{\pi }{2}\] \[\Rightarrow \] \[a=\frac{{{d}^{2}}s}{d{{t}^{2}}}=-\frac{{{\pi }^{2}}}{18}\sin \frac{\pi t}{2}\] At \[t=1s,\] \[a=-\frac{{{\pi }^{2}}}{18}cm/{{s}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
