How Do You Calculate the Radial Component of Acceleration in a Turning Car?

[jonnejon]

Homework Statement

A car speeds up as it turns from traveling due south to heading due east. When exactly halfway around the curve, the car's acceleration is 2.70 m/s^2, 10.0 degrees north of east.

What is the radial component of the acceleration at that point?

Homework Equations

I have some equations in my notes but not sure which one to use.

The Attempt at a Solution

I am trying to study for a test. I got this problem and the answer but I have no clue where to start. Can someone help me?

I am assuming 2.70m/s^2 is the acceleration and I want to find the centrifugal acceleration right? Please help?
A: 2.21m/s^2

 

[kdv]

Homework Statement

A car speeds up as it turns from traveling due south to heading due east. When exactly halfway around the curve, the car's acceleration is 2.70 m/s^2, 10.0 degrees north of east.

What is the radial component of the acceleration at that point?

Homework Equations

I have some equations in my notes but not sure which one to use.

The Attempt at a Solution

I am trying to study for a test. I got this problem and the answer but I have no clue where to start. Can someone help me?

I am assuming 2.70m/s^2 is the acceleration and I want to find the centrifugal acceleration right? Please help?
A: 2.21m/s^2

You have to make a drawing. Draw an arc of a circle showing the motion of the car from due South to due East (so you will have a fourth of a circle). Now consider the point midway along this arc. At that point, draw the acceleration vector in the direction they give you.

Your goal is to find the component of this acceleration vector which points toward the center of the circle.

In general, this would be difficult but here you are in luck because you are at the point midway between due South and due East. Because of this, the radius of the circle will point exactly 45 degrees North of East.

 

[jonnejon]

Thanks. So the formula is: V=Vxcos(x) => A=Acos(x) => V=2.7cos(35)= 2.21m/s^2?

I didn't know you can use the velocity of an axis formula for acceleration also. Please clarify if I did it right. It doesn't look right but I got the answer of 2.21m/s^2.

 

[kdv]

Thanks. So the formula is: V=Vxcos(x) => A=Acos(x) => V=2.7cos(35)= 2.21m/s^2?

I didn't know you can use the velocity of an axis formula for acceleration also. Please clarify if I did it right. It doesn't look right but I got the answer of 2.21m/s^2.

Well, that's the correct equation but you should not use it blindly. It is not that you are using a "velocity of an axis formula", it's simply that you have a vector an dyou are looking for its component along a certain axis.

You should really draw an arc of a circle. The you should draw a line going from the position of the object toward the center of the circle. Now draw the acceleration vector. Do you see that the componen of the acceleration vector along the line going to the center of the circle has acomponent of [itex]a \cos 35 [/itex]? If you do those steps and you see where this equation comes from, you will have understood the problem.

 

[jonnejon]

I drew a diagram of the problem. From the problem I was thinking about using trigonometry because it was a triangle. So I tried an equation from my notes and the closes equation with trigonometry was a v=vxcos(x). I just didn't think that a can be substituted for v. Thanks again. I understand the problem.