A) \[\sin \,t+\frac{1}{2}\cos \,2t\]
B) \[\cos t-\frac{1}{2}\sin \,2t\]
C) \[\sin t-\frac{1}{2}\sin \,2t\]
D) \[\sin t+\frac{1}{2}\sin \,2t\]
Correct Answer: C
Solution :
[c] As we know, \[F=ma\Rightarrow a\propto F\,\,or,\,a\propto \,\,\sin \,t\] \[\Rightarrow \frac{dv}{dt}\propto \sin t\Rightarrow \int\limits_{0}^{0}{dV\propto \int\limits_{0}^{t}{\sin t\,dt}}\] \[V\propto -\cos \,t+1\] \[\int\limits_{0}^{x}{dx}=\int\limits_{0}^{t}{(-\cos t+1)dt}\] \[x=\sin \,t-\frac{1}{2}\sin 2t\]
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