Dynamics Course question

1. Mar 28, 2006

markow202

Im trying to figure out in this question what to do and how to put the formula together.

"A particle travels in a straight line such that for a short time
2s < t < 6s its motion is described by v=(4/a) ft/s, where a is in ft/s^2. If v=6ft/s when t=2s, determine the particle's acceleration when t=3s."

I figured to use a= dv/dt formula but not sure.

If anyone can help...:yuck:

2. Mar 28, 2006

Galileo

Yeah, dv/dt will give you the acceleration. But I`m a bit confused about the question. What is the 'a' in your velocity equation? It has the units of acceleration, but it's at the bottom, so whatever the physical quantity the 4 is must have dimension ft^2/s^3. Did you copy the problem correctly?

As its stands, it looks like the velocity does not even depend on time...

3. Mar 28, 2006

markow202

Im confused aswell about that. I got that equation from the examples in the text.

examples says: "knowing v=f(t) the acceleration is determined from a=dv/dt, since equation relates a,v and t."

When I use this dv/dt formula, which numbers do I divide exactly?

4. Mar 28, 2006

Galileo

Yeah, the acceleration a is the rate of change (i.e. the time derivative) of the velocity. The derivative dv/dt is not simply a ratio between two numbers.

By the way, this really does not belong in the advanced physics threads.

5. Mar 28, 2006

markow202

Not sure where it belonged.....