Acceleration as a Function of Velocity

AI Thread Summary
The discussion revolves around the relationship between acceleration and velocity in a physics problem where velocity is defined as v = 4/a. Participants express confusion regarding the units and the formulation of acceleration, with the correct expression being a = 4/v. There is a consensus that the constant 4 does not have units, simplifying the relationship. The focus is on clarifying how to derive acceleration from the given velocity equation. The conversation highlights the importance of unit consistency in physics equations.
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
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top