1. Oct 9, 2008

### kblue!1

1. The problem statement, all variables and given/known data
A car at the Indianapolis 500 accelerates uniformly from the pit area, going from rest to 250 km/h in a semicircular arc with a radius of 230 m.

Determine the tangential and radial acceleration of the car when it is halfway through the turn, assuming constant tangential acceleration.

r=230m
v1=0 m/s
v2= 250km/h --> 69.44m/s

2. Relevant equations

a=r $$\alpha$$

Vtangent=$$\sqrt{\frac{GM}{r}}$$

3. The attempt at a solution

w1= $$\frac{v1}{r}$$=0

w2=$$\frac{v2}{r}$$=$$\frac{69.44m/s}{230m}$$=0.302rad/s

(.302)2= 02 + 2 ($$\pi$$/2) $$\alpha$$

$$\alpha$$ = .02903 radan/s2

I got this for my TANGENT: (230m)(.02903radan/s2) = 6.68m/s2

and this is my RADIAL:$$\alpha$$ = 0.0290m/s2

2. Oct 10, 2008

### alphysicist

Hi kblue!1,

The speed of .302rad/s corresponds to when the car has moved through the semicircular path, so I don't think the angle is pi/2 here.

The alpha value is the angular acceleration; the radial acceleration that the question asks for is related to the radius and the angular velocity. What formula does it have?

3. Oct 10, 2008

### kblue!1

the formula for radial acceleration is V^2/r

Is it just pi?

4. Oct 10, 2008

### alphysicist

Yes, it would be pi (since the speed of 0.302 rad/s is after the car has moved through an angle of pi).

The radial acceleration is v^2/r like you have; this is also equivalent to

$$a_r=r\ \omega^2$$
so you can find either v or $\omega$ at the halfway mark, whichever you prefer.

5. Oct 10, 2008