Acceleration as a Function of Velocity

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
DeboPGH
Messages
3
Reaction score
0

Homework Statement



When 2<=t<=6
v= 4/a in ft/sec2

v= 6 ft/sec @ t=2s

Find a at t=3s


Homework Equations



a=dv/dt.

The Attempt at a Solution



My integration is horribly wrong and honestly not worth typing. LOL
 
Physics news on Phys.org
DeboPGH said:

Homework Statement



When 2<=t<=6
v= 4/a in ft/sec2

v= 6 ft/sec @ t=2s

Find a at t=3s


Homework Equations



a=dv/dt.

The Attempt at a Solution



My integration is horribly wrong and honestly not worth typing. LOL

You write,

v= 4/a in ft/sec^2

something is goofy here, something is not right?
 
Spinnor said:
You write,

v= 4/a in ft/sec^2

something is goofy here, something is not right?

I'm sorry...acceleration (a) should be in ft/sec^2

so a=4v^-1
 
Now you have acceleration = 4/velocity what are the units of the 4

acceleration * time has units of velocity. Your expression a=4v^-1 still confuses me. Does the number 4 have units? I'm missing something that others might be missing as well.
 
Spinnor said:
Now you have acceleration = 4/velocity what are the units of the 4

acceleration * time has units of velocity. Your expression a=4v^-1 still confuses me. Does the number 4 have units? I'm missing something that others might be missing as well.

No units on 4 it's just a constant?

the original problem is v = 4 divided by acceleration or v = 4/a

I solved for acceleration (a hint by my professor) so acceleration = 4 divided by velocity or a = 4/v