First slide

The two types of acceleration

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 $\vec{v} = (6 .0 m / s) \hat{j}$ . At t = 5.0 s, it moves through (11.0 m, 0) with velocity $\vec{v} = (- 6 .0 m / s) \hat{j}$ . What is its acceleration at t = 2.5 s?

Moderate

A

$(9 .0 m / s^{2}) \hat{j}$
B

$(4 .0 m / s^{2}) \hat{j}$
C

$(- 9 .0 m / s^{2}) \hat{j}$
D

$(- 4 .0 m / s^{2}) \hat{j}$

Solution

The motion of the particle is shown in the following figure:
Radius, $r = \frac{(11 .0 - 3 .0)}{2} = 4 .0 m .$
Magnitude of acceleration,
$= \frac{v^{2}}{r} = \frac{{(6 .0)}^{2}}{4 .0} = 9 .0 m / s^{2} .$
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 get
$\vec{a} = (- 9 .0 m / s^{2}) \hat{j}$

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