Need help on finding physic formulas

[Swordsman318]

I need to figure out 2 formulas for my science fair project:
What is the formula for "velocity as a function of time when an object is under constant acceleration".
What is the formula for "distance as a function of time when an object is under constant acceleration".


I need help on finding these formulas. Please Help!

[deleted-71709]

Velocity, distance and time are all related by a single formula. Velocity is a measure of how far an object can move in a given amount of time while it is not accelerating.

First, let's talk a little about each. We'll use English units of measure.

Time is commonly measured in hours, minutes or seconds.

Distance is often measured in feet or miles.

Velocity is commonly measured using all of the above, such as miles per hour, or feet per minute.

Here is the equation that relates velocity, distance and time, when acceleration is constant:

velocity = distance / time -or- V = D / T

That is: velocity equals distance divided by time.

Using simple algebra, we can describe distance using velocity and time like this:

distance = velocity x time -or D = V x T

I hope that helps.

Ed Neu
Buffalo, MN

[deleted-71588]

The previous response is incomplete...

With constant acceleration, velocity will be increasing with time: V(t) = V(t=0) + a t [Notationally, velocity V is a function of time V(t) that starts at some initial velocity V(t=0)]
Whoa, where did that come from? Calculus. a(t) = dV(t)/dt - Acceleration (at some time t) is the change of velocity (at that time t) with respect to time. Because acceleration was given to be a constant, a(t) is constant for all time so it is not a function of time, so a = dV/dt for all t. What this says is that the change in velocity with respect to time is constant and is in fact equal to the acceleration. In this calculus notation d../d.. is indicates the change (difference, differential, derivative) of something (in this case V, velocity function) with respect to something (in this case t, time variable). Can you figure this out without calculus? IMO: Not without memorizing a bunch of formulas for special cases. Even if you don't understand all of calculus, you can understand some of the notation enough to convert it back to a word description of a relationship or problem statement. Because the previous responder didn't look at this as a calculus problem or use functional notation, they didn't catch that velocity was a function of time and not a constant.

If we let V(t=0) be represented by Vo, the initial velocity, the V(t) = Vo + a t, or velocity starts out at the initial velocity and linearly increases by the a (acceleation constant) times the amount of time elapsed, t.

Since I've used "d" for the calculus notation, I don't want to use it for distance, so I'm going to use X instead.

ASSUMPTION: X(t=0) = 0, in other words, lets put the initial starting point on the origin of our distance scale.

X(t) = Vo t + (t/2)(V(t) - Vo)
Whoa, where did this come from. For me, it came from calculus; however, you don't need calculus for this one. A bit of geometry will do.
If the velocity were constant, then you would get X(t) = Vo t, the initial velocity times the elapsed time, but the velocity is not constant.
Because the velocity is a linear function of time, if we subtract off the initial velocity and plot the result, it will look like a line.
If we want the area under this line, it will represent the area of a right triangle. The time side of the triangle is t, the velocity side is (V(t) -Vo), and the 1/2 comes from the formula for the area of a right triangle. You might also recognize that Vo + (V(t) - Vo)/2 = (V(t) + Vo) / 2 which is average velocity over the period from 0 to t. Again, because velocity is a linear function of time, you can use the average velocity in the distance equation:
X = Va t or distance equals average velocity time time, where Va = (V(t) - V(t=0))/2 = (V(t) -Vo)/2