What is the formula for final velocity that has time in it?

AI Thread Summary
The discussion focuses on finding the formula for final velocity that incorporates time. The key equation provided is v = u + at, where v is the final velocity, u is the initial velocity, a is acceleration, and t is time. Another equation mentioned is v^2 = u^2 + 2as, but it does not directly include time. A more complex formulation is also presented, but it ultimately simplifies back to the primary equation. This information is crucial for students preparing for physics exams.
∞infinite∞
Messages
1
Reaction score
0
Hi,

I'm studying for my final exam in Physics tomorrow (doing a review packet) but I keep forgetting the equation/formula needed to find final velocity (not the one with the distance, but the one with time). I don't have most of my notes as my teacher keeps folders of our work and notes at school... and I can't find it on Google.

Please help ASAP it's 11:20 PM here, my eyes are killing me and I want to get an A on the exam...

Thanks in advance.


PS: I think it starts with "vf^2= ... at" ?
 
Physics news on Phys.org
There isn't one for v^2 involving t, the equations for the final velocity are

v = u + at

and

v^2 = u^2 + 2as

whereu is the initial velocity and v is the final velocity.
 
Although Vagn is correct, you could say that

{V_f}^2 = {V_i}^2 + 2a \left( \frac{V_i + V_f}{2} \cdot t \right)


but, ultimately, this simplifies to:

V_f = V_i + at
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top