# Homework Help: Accelleration Through a Curve

1. Sep 14, 2006

### NotMrX

If a car gradually changes a direction of 90 degrees at constant speed of 200 m/s over a time period of 20 seconds, then what is the accelleration?

I don't know if the way I worked it was correct. I think the only assumption is that the acceleration is constant.

$$a_x=\frac{v_xf-v_xi}{t}=\frac{0-200}{20}=-10$$
$$a_y=\frac{v_yf-v_yi}{t}=\frac{200-0}{20}=10$$
$$a=\sqrt{a_x^2+a_y^2}=\sqrt{(-10)^2+10^2}=10\sqrt{2}$$

Does this seem right to you?

2. Sep 14, 2006

### Andrew Mason

No. Assume it is a curve of constant radius. Work out the centripetal acceleration. Is there any tangential acceleration here? Take the vector sum of both accelerations.

AM

3. Sep 14, 2006

### NotMrX

How would someone do that if the radius isn't given? I understand that
$$a_c=\frac{v^2}{r}$$
but if there is no r value it seems hard to figure out the centripetal acceleration?

4. Sep 14, 2006

### Andrew Mason

Use $\theta = \omega t$ and $v = \omega R$

AM

5. Sep 14, 2006

### NotMrX

$$a_c=\frac{v^2}{r}=\frac{v^2}{v/\omega}=v*\omega=v*\frac{\theta}{t}=200*\frac{\pi/2}{20}=5\pi$$

Thanks for the help. How come the other way didn't work?

6. Sep 14, 2006

### Andrew Mason

Because $\sqrt{2} \ne \pi/2$. Close but not the same.

You will notice that the components do not change uniformly with time - it depends on the angle. You are taking the sum of the time average of each component over a quarter turn. Averages over time will be different than the instantaneous value unless the rate of change is uniform over time.

AM