Question

A particle moves along a circle of radius 'R' with speed varies as V=apt, Where ao is a positive constant Then the angle between
the velocity vector and the acceleration vector of the particle when it has covered one fourth of the circle is
IA​

Answer 1

Given : A particle moves along a circle of radius 'R' with speed varies as V =  a₀t  

To find : angle between the velocity vector and the acceleration vector of the particle when it has covered one fourth of the circle

Solution:

V = a₀t

at t = 0  V = 0

=> u = 0

a = dV/dt = d(a₀t)/dt  = a₀

using v² - u² = 2aS

=>  v²  - 0 = 2a₀S

S= (1/4)2πR  = πR/2

=>  v²  = 2a₀ πR/2

=>   v²  =  a₀ πR

=> v²  / R = πa₀  

=> $a_{n} =$ πa₀  

$a_{t} = a_{0}$

the angle between the velocity vector and the acceleration vector   = α

Tan α  =  $\frac{a_{n}}{a_{t}}$

=> Tan α = πa₀   / a₀

=>  Tan α = π

=> α =  Tan⁻¹(π)

Tan⁻¹(π)  is the angle between the velocity vector and the acceleration vector of the particle when it has covered one fourth of the circle

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