# Car Radial and Tangential Acceleration

1. Sep 30, 2007

### JeYo

A car speeds up as it turns from travelling due South to due East. Exactly halfway around the curve, the car's acceleration is 3.40m/s^2, 30.0º north of east.

We are supposed to find the radial and tangential components of the acceleration at the halfway point.

I thought that I could break the acceleration vector into its parallel and perpindicular components but I am not entirely sure of where the acceleration vector is located. I really would love a push in the direction of the answer because I know with a little push I can make most of the way.

2. Oct 1, 2007

### JeYo

Alright, I figured out the radial component of the acceleration but I am still somewhat unable to figure out the tangential component of the acceleration.

3. Oct 1, 2007

### learningphysics

Draw a picture... suppose it is driving on a circular path... going counterclockwise... it is initially at the west end of the circle going south... soon it will be south end of the circle going east...

so it is halfway between these two points... so it is south 45 degrees west. What is the direction of the centripetal acceleration?

what is the component of the given accleeration along this direction (that's the radial acceleration)... what is the component of the given acceleration perpendicular to this direction (that's the tangential acceleration).

4. Oct 1, 2007

### JeYo

I have a pretty good pictorial representation of the problem and I know in which directions and from whence the three vectors point. But the connection between these and the value has escaped me. Because I do not know how to find the numerical quantity that represents the tangential acceleration. I thought that maybe the a = (a_r^2 + a_t^2)^.5 would work if I used the value I found for a_r = 3.28m/s and the given value 3.40m/s for a. But it does not seem to work.

5. Oct 1, 2007

### learningphysics

What you did should work... perhaps it's the decimal places.

The centripetal acceleration is in the direction E 45 degrees N. The net acceleration is E 30 degrees N.

you want to divide the net acceleration into two components... one along the line E 45 degrees N... and another perpendicular to that.

So you have a right triangle with an angle of 15 degrees.

a*cos(15) gives radial acceleration
a*sin(15) gives tangentail acceleration.

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