How Do You Calculate Total Acceleration of a Particle on a Circular Path?

[Sheen91]

Homework Statement

A particle starts from rest at t=0 s.
It moves along a circular path of radius 18m and has an acceleration component along its path of 4.3m/s^2.

What is the magnitude of the acceleration when t=3 s?

Homework Equations

N/A

The Attempt at a Solution

I am not sure how to start. If the particle is accelerating at 4.3m/s^2 at t=0s then will it not be at that same acceleration at t=3 s?

I am sure it is not a trick question, as I know the acceleration will start out along the path of the particle and then the acceleration will point inward toward the center of the circle path.

Any and all help will be much appreciated.

Cheers

 

[rl.bhat]

Find the velocity of the particle after 3 seconds by using relevant kinematic equation. Then find the centripetal acceleration a that instant. The net acceleration of the particle is the resultant of the tangential and the radial acceleration.

 

[Sheen91]

Thank you so so much. Trying that now.

so

magnitude of acceleration = v^2/r

Acceleration is 4.3m/s^2 so at t=3 s

v = 4.3 x 3 = 12.9m/s

mag of acc = 12.9^2/18 = 9.245m/s^2

Does that seem right? Just want to double check before submitting :D

 

[rl.bhat]

Thank you so so much. Trying that now.

so

magnitude of acceleration = v^2/r

Acceleration is 4.3m/s^2 so at t=3 s

v = 4.3 x 3 = 12.9m/s

mag of acc = 12.9^2/18 = 9.245m/s^2

Does that seem right? Just want to double check before submitting :D

You are right. Now find the resultant of 4.3 m/s^2 and 9.245 m/s^2

 

[Sheen91]

9.245 m/s^2 is normal to 4.3 m/s^2

so you would just do this?

$$\sqrt{4.3^2 + 9.245^2}$$ = 10.196 m/s^2

 

[rl.bhat]

9.245 m/s^2 is normal to 4.3 m/s^2

so you would just do this?

$$\sqrt{4.3^2 + 9.245^2}$$ = 10.196 m/s^2

You are right.