Instantaneous Velocity

heartofaragorn

1. The problem statement, all variables and given/known data
A workman drops a wrench over the side of a high-rise building. Someone looking out their window sees the wrench appear at 11:00 a.m. Another person in an office 25.0 m below the first observer sees the wrench appear exactly 1.00 second after 11:00 AM.

Calculate the instantaneous velocity of the wrench when it passes the second observer, calculate the total distance the wrench has fallen from the point when it passes the second observer, and calculate the time when the wrench was dropped.


2. Relevant equations
Instantaneous velocity = change in velocity / change in time
v = v initial + accleration X time
x - x initial = v initial X t + 1/2 acceleration X time squared


3. The attempt at a solution
It seemed to me that the instantaneous velocity would be 25 m/s squared, but since that answer is too simple, I'm pretty sure it's wrong. I looked through the kinematic formulas but couldn't find one that I had enough information for. For instance, how do I calculate this using only on measurement (the 1.00 second it took to go 25 m?). Please help!
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

robphy

Check this statement.
What are the units of velocity?

nrqed

You know the acceleration, right? You know the distance travelled and the time interval. You can find both the initial velocity (when it passes the first window) and the final velocity. See below

This is incorrect. This equation gives the average acceleration, not the instantaneous velocity
That's it. Use the second equation to find the initial velocity. Then use the first equation to find the final velocity.
(here, the initial point corresponds to the wrench passing the first window and the final point is when the wrench passes the second window.

Once you have that information, use again the equations but now the initial point being when the wrench is dropped (with an initial velocity of zero) and the final point being when it passes the first window (say). Then the unknowns will be the time and the distance, which you can solve for using the same two equations

Doc Al

The average velocity would be 25 m/s (not squared!) during that 1 second interval. But that's not the instantaneous velocity.

How must the instantaneous velocities at the beginning and end of that interval differ? Hint: What's the acceleration?

heartofaragorn

Okay, thank you for the help, everyone! That definitely helped me out, so much so, that I'm about to post another problem!

Feldoh

You give the formula to find the average velocity.
[tex]\lim_{h\rightarrow 0} \frac{s(a+h)+s(a)}{h}[/tex] where t=a is instantaneous velocity^^