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→$$=(6.0m/s)j^$$. At t $$=$$ $$5.0$$ s, it moves through (11.0$$ m, $$0) with velocity v→$$=(-6.0m/s)j^$$. 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→$$=(6.0m/s)j^$$. At t $$=$$ $$5.0$$ s, it moves through (11.0$$ m, $$0) with velocity v→$$=(-6.0m/s)j^$$. 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, $$r=(11.0$$−$$3.0)2=4.0mMagnitude$$ of acceleration,$$a=v2r=(6.0)24.0=9.0m/s2at$$ 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 centre, We geta→$$=$$−$$9.0m/s2j^$$

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

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→=(6.0m/s)j^. At t = 5.0 s, it moves through (11.0 m, 0) with velocity v→=(-6.0m/s)j^. What is its acceleration at t = 2.5 s?

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material