Question

a particle is moving in a circular path of radius R =1 m , its speed varies with time(t) as v=3t m/s find its total acceleration at t = 1 second​

Answer 1

Explanation:

Given,

Radius of the circular path, r=1 m

Speed, v=3t m/s

Total acceleration = square root of { (Normal acceleration)² +(Tangential acceleration)²}

Normal acceleration, A¹¹ =v²/r where r=radius = 1m

A¹¹ =(3t m/s)² /1 m

A¹¹=9t² m/s²

at t = 1 second , we have

A¹¹ =9 m/s²

And

Tangential acceleration ,A¹² = dv/dt =d(3t)/dt

A¹² = 3 m/s²

Now ,

Total acceleration = square root of { (Normal acceleration)² +(Tangential acceleration)²}

Total acceleration = square root of { (9 m/s²)² +(3 m/s²)²}

Total acceleration = square root of {81 (m/s²)² +9(m/s²)²}

Total acceleration = square root of { 90(m/s²)² }

Total acceleration = 30 m/s² solved :::::: Hope you like it