# Centripedetal Acceleration with a decelerating car.

1. Sep 8, 2008

### CaptainSFS

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

A car moving at a speed of 30 m/s enters a curve that describes a quarter turn of radius 114 m. The driver gently applies the brakes, giving a constant tangential deceleration of magnitude 1.2 m/s2.

a) Just before emerging from the turn, what is the magnitude of the car's acceleration?

b) At that same moment, what is the angle q between the velocity vector and the acceleration vector? Please enter your answer in degrees.

2. Relevant equations
Kinematics
i tried: a=(v(squared)/R)
or maybe another sort of similar formula like a=omega(squared)*R

3. The attempt at a solution

a) I used a=(v(squared)/R). I tried finding the velocity at the end of the turn, I tried using 37.01661958 (the new vector formed by 30m/s and final v (21.68479018). I used 114m for the radius. There's something that i dont understand about these circular formulas and how I am supposed to solve them.

b) I believe I am supposed to use my new vector (37.01661958) in relation to the acceleration (which i believe is pointing at the center of the circle), but I can't remember how to calculate that angle.

Any help is very much appreciated!
thanks,
SFS

2. Sep 8, 2008

### LowlyPion

So what was the direction of the acceleration of the car when it began its turn?

It was directed toward what if you are describing it as V2/R ?

Now if you are using that to describe the acceleration on the car in the turn, what magnitude will acceleration be then when it is slowed by the tangential deceleration? And how is it directed?

3. Sep 8, 2008

### CaptainSFS

Um.. I guess the acceleration of the car when it began is facing behind the car. But because it's turning, isn't there an acceleration facing the center of the circle too (if the turn was a complete circle)?

I'm also not sure how to calculate that magnitude of the slowed acceleration, and I am still unsure how it is directed.

4. Sep 8, 2008

### LowlyPion

The car until it hits the turn is not undergoing any acceleration or deceleration. It is 30 m/s. When it begins its turn on the 114 m radius turn it experiences the centripetal acceleration of V2 / R. And as you point out that is radially directed.

At the 1/4 turn of the turn then where would that acceleration still be pointing? It's magnitude is smaller because of the tangential deceleration. But the direction of the centripetal acceleration is still where?

When you add in the tangential deceleration vector, to the centripetal acceleration vector ...

5. Sep 8, 2008

### CaptainSFS

okay, i think it is slightly making more sense. The centripetal is always facing to the inside of the turn (i.e. center of the cirlce). So at the end of the 1/4 turn its still facing the center, but at a different angle. And the tangential acceleration is the tangent of the curve at the end of the curve?

I am still really clueless in calculating these two accelerations, or maybe just the centripetal acceleration because the tangential is given?, because when I add the two numbers that i come up with, it still isn't the correct answer.

6. Sep 8, 2008

### LowlyPion

Yes that's it.

Centripetal acceleration is v2/r
The tangential acceleration is -1.2 m/s2 as given.

These are at right angles but they resolve into a resultant acceleration vector.

You can calculate the v2/r from the Velocity that it slowed to over the course of the turn. (You already calculated that.)

7. Sep 8, 2008

### rgalvan2

I am also having difficultly with this problem. I'm confused on how or even where to start. Help please!! This is due tomorrow!

Last edited: Sep 8, 2008
8. Sep 8, 2008

### CaptainSFS

Oh, thanks! I figured i had to calculate a resultant vector, but i thought it was for the initial and final velocity. The resultant acceleration vector i got worked. So thanks again for helping explain this to me, it makes much more sense.

peace.