A particle is moving on a circular path of radius 1m. It starts from rest, its tangential acceleration is 3t. The time after which the acceleration of the particle makes an angle of 300 with centripetal acceleration is (n)2/3s. Find the value of n.

acos300=at and a sin 300=at ∴tan300=atat=3tat at=3t or v21=3t ∴v2=3tBut at=dvdt=3t or ∫0vdv=3∫0ttdt or v = 32t2 ∴34t4=3t or t3=4Given t=22/3s∴n=2

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A particle is moving on a circular path of radius 1m. It starts from rest, its tangential acceleration is $\sqrt{3} t$ . The time after which the acceleration of the particle makes an angle of 30⁰ with centripetal acceleration is ${(n)}^{2 / 3} s$ . Find the value of n.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 2.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$a \cos 30^{0} = a_{t} {and a sin 30}^{0} = a_{t} ∴ \tan 30^{0} = \frac{a_{t}}{a_{t}} = \frac{\sqrt{3} t}{a_{t}} a_{t} = 3 t o r \frac{v^{2}}{1} = 3 t ∴ v^{2} = 3 t$

But $a_{t} = \frac{d v}{d t} = \sqrt{3} t$ or $\int_{0}^{v} d v = \sqrt{3} \int_{0}^{t} t d t or v = \frac{\sqrt{3}}{2} t^{2}$
$∴ \frac{3}{4} t^{4} = 3 t o r t^{3} = 4$
Given
$\begin{array}{l} t = 2^{2 / 3} s \\ ∴ n = 2 \end{array}$