question_answer

The acceleration of a particle, starting from rest, varies with time according to relation ;\[\text{a}=-\text{r}{{\omega }^{\text{2}}}\text{sin}\omega \text{t}\]. Find the displacement of this particle at a time t.

Answer:

                Acceleration, \[\text{a}=\frac{dv}{dt}=\frac{dv}{dx}\times \frac{dx}{dt}=\upsilon \frac{dv}{dx}\] Given,\[{{\upsilon }^{\text{2}}}=\left( \text{18}0-\text{16x} \right)\]; Differentiating it w.r.t. x, we have \[\text{2}\upsilon \frac{dv}{dx}=-\text{16}\] So, acceleration, \[a=\upsilon \frac{dv}{dx}=-\frac{16}{2}=-\text{8m}/{{\text{s}}^{\text{2}}}\]