I am posting here being fully aware that this thread is more than 8 years old and that the OP has not been seen for almost 7 years. I believe that unanswered threads in the "Related Threads ..." section of the Introductory Physics Homework threads are of no help to anyone. For that reason I am posting a complete solution for the benefit of all. This old thread is related to the current thread
https://www.physicsforums.com/threads/instantaneous-acceleration-from-coordinates.809726/.
Lest it be thought that I am violating forum rules by doing so, I point out that the span of eight years most likely exceeds the statute of limitations from the date of the original post. I base this conclusion on my assessment that such violation of PF rules can be no more egregious than the following felonies that have a statute of limitations of only five years in the state of Texas where I reside. (
https://brettpodolsky.com/criminal-...e-of-limitations-expired-for-your-texas-crime)
Under Tex. Code Crim. Proc. Ann. Art. 12.01(4), specific felony offenses have a five-year statute of limitations from the date on which the crime was allegedly committed, including:
- Robbery
- Theft
- Kidnapping
- Burglary
- Injury to a disabled or elderly individual (unless punishable as a first-degree felony under Texas Penal Code Section 22.04)
- Endangering/abandoning a child
Now onto the solution.
Using conventional Cartesian coordinates, we write the initial and final velocities as ##\vec v_i=\omega R~\hat x~;~\vec v_f=-\omega R~\hat y.## The average acceleration is$$\vec a_{avg.}=\frac{\vec v_f-\vec v_i}{\Delta t}=-\frac{\omega R} {\Delta t}(\hat x+
\hat y).$$ With ##\omega=\dfrac{2\pi}{60} \mathrm{s^{-1}}## and ##R=0.3~ \mathrm{m}##, the expression evaluates to ##\vec a_{avg.}=-2.1\times 10^{-3}(\mathrm{m/s^2})(\hat x+\hat y).## OP's answer
Error said:
is correct.
OP did not realize that the instantaneous acceleration at the 12:00 position is centripetal and given by ##\vec a=\omega^2 R~(-\hat y)=-3.3\times 10^{-3}(\mathrm{m/s^2})~\hat y.##