JEE PHYSICS MCQ: MOTION IN PLANE

1 Medium

1 of 1

Application +4 | -1

A particle is moving in the

x-y

plane and its coordinates at any time

t

are given by

\begin{aligned} & x=5 \cos \omega t \\ & y=5 \sin \omega t \end{aligned}

Where

(\omega)=\frac{\pi}{4} \mathrm{rad} / \mathrm{s}

. The direction of force it experiences at

t=3 \mathrm{~s}

is.

View answer and explanation

Correct option: B

\hat{i}-\hat{j}

Option B (

\hat{i}-\hat{j}

) is correct because it represents the direction of the net acceleration (and therefore the force) at t=3s, considering both x and y components.

More Questions from MOTION IN PLANE

Practice All Chapter MCQs →

\mathrm{\vec P} = \mathrm{\vec A} + \mathrm{\vec B}\,\,\mathrm{and}\,P = A + B.

\mathrm{\vec A}\,and\,\mathrm{\vec B}

0^{\circ}

90^{\circ}

\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0

(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b})=0

\mathbf{a}

\mathbf{b}

\theta

\left(90^{\circ}-\theta\right)

39.2 \mathrm{~ms}^{-1}

50 \mathrm{~m}

{g}=9.8 \mathrm{~ms}^{-2}