A particle undergoes uniform circular motion on a horizontal xy plane. At time t $$=$$ $$0$$, it moves through coordinates (3.0$$ m, $$0) with velocity v→$$=6.0$$ $$m/sj^$$. At t $$=$$ $$5.0$$ s, it moves through (11.0$$ m, $$0) with velocity v→$$=$$−$$6.0$$ $$m/sj^$$. What is its acceleration at t $$=$$ $$2.5$$ s?

Question

A particle undergoes uniform circular motion on a horizontal xy plane. At time t $$=$$ $$0$$, it moves through coordinates (3.0$$ m, $$0) with velocity v→$$=6.0$$  $$m/sj^$$. At t $$=$$ $$5.0$$ s, it moves through (11.0$$ m, $$0) with velocity v→$$=$$−$$6.0$$ $$m/sj^$$. What is its acceleration at t $$=$$ $$2.5$$ s?

Infinity Learn · Accepted Answer

The motion of the particle is shown in the following figure:Radius, $$r=11.0$$−$$3.02=4.0$$ m.Magnitude of acceleration, $$=v2r=6.024.0=9.0$$  $$m/s2$$.at t $$=$$ $$2.5$$ s, the particle has completed $$one-fourth$$ of its period and its velocity is parallel to $$x-axis$$. Since the centripetal acceleration is directed towards to the center. We geta→$$=$$−$$9.0$$ $$m/s2j^$$

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A particle undergoes uniform circular motion on a horizontal xy plane. At time t = 0, it moves through coordinates (3.0 m, 0) with velocity v→=6.0 m/sj^. At t = 5.0 s, it moves through (11.0 m, 0) with velocity v→=−6.0 m/sj^. What is its acceleration at t = 2.5 s?

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