# Centripetal Acceleration formula derivation

• I
Hello,
This is a very basic question. I am sure I am doing something wrong in the derivation as shown in the picture. But I am not able to find out where I am doing it wrong. It would be very helpful if you can pls. let me know what I am doing wrong here.
Thanks

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• Centrepetal.jpg
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## Answers and Replies

Doc Al
Mentor
You calculate the distance traveled in Δt. Why? And then you set that distance equal to the change in velocity? Why again?

• SamitC
You calculate the distance traveled in Δt. Why? And then you set that distance equal to the change in velocity? Why again?
Because in Δt, the velocity changes from v0 to v. And for small change in time this can be equal to the distance traveled because of the triangle similarities. That is what the assumption is behind the effort. Can you pls. elaborate where you find something wrong or missing ?

Actually,the assumption that for a small time interval, the change in velocity is equal to the distance travelled is wrong.

• SamitC
Doc Al
Mentor
Because in Δt, the velocity changes from v0 to v. And for small change in time this can be equal to the distance traveled because of the triangle similarities. That is what the assumption is behind the effort. Can you pls. elaborate where you find something wrong or missing ?
A change in velocity cannot equal a distance. You can use similar triangles: ##\frac{\Delta v}{v} = \frac{\Delta x}{r}##. Use that and you can easily derive the formula.

• SamitC
A change in velocity cannot equal a distance. You can use similar triangles: ##\frac{\Delta v}{v} = \frac{\Delta x}{r}##. Use that and you can easily derive the formula.
I got it now. It has to be the ratio...Thanks a lot