# Magitude of accelaration

rasikan

## 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 don't know how to slove with this equation

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?

rasikan
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?

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

rasikan
the two components are ax and ay

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?

rasikan
coz that is a constant acceleration?

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?

rasikan
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??

rasikan
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?

Homework Helper
towards its center

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

rasikan
u mean my answer is right??