What Do These Constant Velocity and Acceleration Formulas Mean?

[COCoNuT]

can someone tell me what these formulas do?

constant velocity(when velocity is constant, which is 0 right?, you use these formulas)...

1.) a(t) = 0 <== when constant velocity, a(t) = 0 right?
2.) v(t) = Vi(what is v(t)? vi = initial)
3.) x(t) = Xi + ViT (what is this formula used for? what is x(t)?i know that Vi= initial velocity, T= time) is x(0) equal to distance and height.

constant acceleration:
4.) a(t) = a(0) (<--- what does this mean?)
5.) v(t)= Vi + at (<--- what is v(t), and is this used to find the final velocity?)
6.) x(t) = Xi + ViT + 1/2at^2 (<--- what is this formula used for? what is it trying to find? what is x(t)? is x(0) equal to distance and height?)

and this formula, which i don't know what it's used for at all, but i know it's important:
Vf^2 = Vi^2 +2ad (d= distance traveled)

sorry for all the questions, but i think if i understand these formulas more, i would be able to do my physics homework easier, because i ALWAYS have trouble trying to find which formula to use.

 

[HallsofIvy]

can someone tell me what these formulas do?

constant velocity(when velocity is constant, which is 0 right?, you use these formulas)...

"1.) a(t) = 0 <== when constant velocity, a(t) = 0 right?"
Assuming that "a(t)" is an acceleration function (you didn't tell us that!), accleration is defined as rate of change of velocity so, yes, it velocity doesn't change, the acceleration is 0: a(t)= 0.

"2.) v(t) = Vi(what is v(t)? vi = initial)"
Are you still assuming that velocity is a constant? If that is true, then yes, velocity at all times, v(t), is that constant and, in particular is the same as the initial velocity: v(t)= v_i, a constant function.

"3.) x(t) = Xi + ViT (what is this formula used for? what is x(t)?iknow that Vi= initial velocity, T= time) is x(0) equal to distance and height."
If your constant velocity is v_i then the distance moved in time T is v_i times T. If your intitial distance from some 0 point was x_i then your new distance will be x_i plus the distance moved: x(t)= x_i+ v_iT.

"constant acceleration:
4.) a(t) = a(0) (<--- what does this mean?)"
Is it the word "constant" you are having trouble with? a(t) is the acceleration at time t. a(0) is the acceration when t= 0. Since the acceleration is constant, it doesn't change- no matter what t is, the acceleration is still the same as it was at t= 0: a(t)= a(0).

"5.) v(t)= Vi + at (<--- what is v(t), and is this used to find the final velocity?)"
v(t) is the velocity at time t. Yes, it can be used to find the final velocity- it can be used to find the velocity at any time. Acceleration is defined as the rate of change of velocity. It is defined by a= (change in v)/t so, multiplying on both sides by t, change in v= at. Add that change to the initial velocity: v(t)= v_i+ change in v= v_i+ at.

"6.) x(t) = Xi + ViT + 1/2at^2 (<--- what is this formula used for? what is it trying to find? what is x(t)? is x(0) equal to distance and height?)"
x(t) is position of an object (which might be "height" if the motion is vertical) at time t. x(0) is the position of the object when t= 0. It is the distance from whatever point was chosen as the "reference". It can be height if you are measuring vertically. As long as the acceleration is constant, the average velocity between time 0 and t is v_i+ (1/2)at and you multiply that by t to get the distance moved. Add that to initial distance (from the reference point) and you get the new distance.

"and this formula, which i don't know what it's used for at all, but i know it's important:
Vf^2 = Vi^2 +2ad (d= distance traveled)"

I presume V_f is the "final velocity", V_i is the "initial" or starting velocity. The advantage of this is that you can calculate the final velocity if you know the distance but not the time.

For example if the constant acceleration is 2 m/sec<sup>2, v_i is 1 m/s and t= 4 seconds, the final velocity, v_f will be 1+ 2(4)= 9 m/s. The distance moved will be (using d= v_it+ (1/2)at2) d= 1(4)+ (1/2)(2)(16)= 4+ 16= 20.
V_f2= 92= 81 m2/sec2
V_i2= 12= 1 m2/sec2
2ad= 2(2)(20)= 80
The formula would give 81= 1+ 80 (m2/sec2)which is certainly true.

If we were told that the initial velocity was 1 m/s, acceleration 2 m/sec2, but that the acceleration lasted for 20 m rather than 4 seconds, we would calculate v_f2= 1+ 80= 81 and so find that the final velocity was 9 m/s without needing to know the time.</sup>

 

[COCoNuT]

thank you sooo much, everything makes sense now