The position vector of a particle $\to R$ as a function of time is given by $\to R = 4 sin (2 π t)^i + 4 cos (2 π t)^j$ , where $R$ is in meters, $t$ is in seconds and $ˆ i$ and $ˆ j$ denote unit vectors along 'x' and 'y'-directions respectively. Which one of the following statements is wrong for the motion of particle?

Path of the particle is a circle of radius

4

m

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Acceleration vector is along

- \to R

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Magnitude of acceleration vector is

\frac{v^{2}}{R}

, where

v

is the velocity of particle

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Magnitude of the velocity of particle is

8

m s^{- 2}

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Solution

The correct option is B Acceleration vector is along $- \to R$
$\to R = 4 sin (2 π t) ˆ i + 4 cos (2 π t) ˆ j T h i s i s t h e v e c t o r e q u a t i o n v = \frac{d R}{d t} = 8 π cos (2 π t) ˆ i - 8 π sin (2 π t) ˆ j a = \frac{d V}{d t} = - 16 π^{2} sin (2 π t) ˆ i - 16 π^{2} cos (2 π t) ˆ j a = - 4 π^{2} (4 sin (2 π t) ˆ i + cos (2 π t) ˆ j) = - 4 π^{2} (\to R) H e n c e, t h e a c c e l e r a t i o n v e c t o r i s a l o n g - \to R$

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Q. The position vector of a particle

\to R

as a function of time is given by

\to R = 4 sin (2 π t)^i + 4 cos (2 π t)^j

Where R is in meters, t is in seconds and

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