Magitude of accelaration

[rasikan]

1. The problem statement, all variables and given/known data
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?



2. Relevant equations
magitude= V^2/R




3. The attempt at a solution

But i dont know how to slove with this equation

[Dick]

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?

[Dick]

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

[rasikan]

the two components are ax and ay

[Dick]

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?

[Dick]

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?

[Dick]

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??

[Dick]

Yeah. Does that surprise you? If you're clear on the two components of the acceleration then I think you can go on to the next problem...

[rasikan]

yaa i got it now dick thanks for you help