AZING!Acceleration Through a Curve: How to Calculate and Understand It

[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?

 

[Andrew Mason]

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?

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

 

[NotMrX]

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

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?

 

[Andrew Mason]

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?

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

AM

 

[NotMrX]

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

AM

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

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

 

[Andrew Mason]

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

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

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