# Magitude of accelaration

## Homework Statement

A particle moves along a circular path having a radius of 2.0 m. At an instant when the speed of the particle is equal to 3.0 m/s and changing at the rate of 5.0 m/s2, what is the magnitude of the total acceleration of the particle?

magitude= V^2/R

## The Attempt at a Solution

But i dont know how to slove with this equation

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Dick
Homework Helper
Acceleration is a vector. There will be a component tangent to the circle and a normal component. Which one is given to you in the problem statement and which do you have to find using your equation?

Dick,
v=3m/s
a=5 m/s^2

I guess I can find the magnitude when v=3 m/s . is that
right?

Acceleration is a vector. There will be a component tangent to the circle and a normal component. Which one is given to you in the problem statement and which do you have to find using your equation?

Dick
Homework Helper
Ok, so what are the two components of the acceleration.

the two components are ax and ay

Dick
Homework Helper
The components are normal to the circle and tangent to the circle, so x,y coordinates are probably not appropriate. I would say the tangent component is 5 m/s^2. Why would I say that?

coz that is a constant acceleration?

Dick
Homework Helper
No, because they told me. If an object is moving on a circle and it's speed is changing at 5 m/s^2, in what direction is that acceleration pointed?

this is my calculation

a=Sq root of (at)^2+(ar)^2

at= 5m/s^2
ar=v^2/r= 9/2=4.5
a=sq root(5^2+4.5^2)
=6.7 m/s^2

is this is right??

towards its center
No, because they told me. If an object is moving on a circle and it's speed is changing at 5 m/s^2, in what direction is that acceleration pointed?

Dick
Homework Helper
towards its center
Your calculation is fine. What does "towards its center" mean? You labelled it as at in your calculation?

u mean my answer is right??

Dick