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Some basic equations

  1. Oct 3, 2004 #1
    I'm trying to teach myself physics out of a textbook I recently bought. Unfortunately, it's doing a poor job of explaining the first couple of equations in the book. Can you explain them and their uses to me? Even though I know their basic meanings, please don't skip over any of the basic concepts as I'm already confused enough. Thanks.

    1. X = VT
    2. a = delta V / delta T
    3. X(t) = Xi + ViT + 1/2at^2
    4. Vf2 = vi2 + 2ax

    By the way, are there any commands for the different symbols? (Delta, exponents, etc.)
  2. jcsd
  3. Oct 3, 2004 #2
    Some help

    Well I'm not sure about # 3 and 4, so i will leave thoes for someone else to explain.

    First here are some common letter choices:

    x is position or distance.

    v is velocity. [how fast the position is changing]
    It can also be written as dx/dt or delta x/delta t. [meaning velocity is a small change in position per small change in time]

    a is acceleration. [ how fast the velocity is changing]
    It can also be written as dv/dt or delta v/delta t. [meaning acceleration is a small change in velocity per small change in time]

    x(t) is position as a function of time
    [meaning the value of x at some time t. example: x(3)= 6, means x=6 when t=3]
    Velocity and acceleration can also be written as a function of time in the same way.

    Next here is some help on your question:

    X=VT [means the position is the velocity multiplied by the time.]
    {So your position will be X after you move at a speed V for an amount of time T}

    a= delta V/delta t [is the definition of acceleration. see above]

    If that didn't make sense I can go into more detail.
  4. Oct 3, 2004 #3


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    #3 and #4 are equations valid only for constant-acceleration motion.
    #3 gives the position as a function of time.
    #4 is a useful equation for the velocity-squared (obtained from #3 and V(t)=Vi+at)

    #2 is an expression for average-acceleration (which is generally not the same as acceleration).
    #1 is probably "the position as a function of time FOR CONSTANT VELOCITY ONLY" (a special case of #3)
  5. Oct 3, 2004 #4
    1. X is this one is "the change in x" in english, thats distance. Its simply saying that the distance traveled is equal to the speed your going times the amount of time you traveled. This equation comes up a lot in various physics problems and should pretty much be automatic in implementation after spending any sort of time solving physics problems.

    2. a in this equation stands for acceleration, delta V is "change in velocity", delta T is "change in time". This is the definition of acceleration.

    3. final position = initial position + initial velocity + 1/2 acceleration * time squared. This equation is used a lot in mechanics and should be known by heart.

    4. I am not sure about the 4th one. It looks an awful lot like Vxf^2 = Vxi^2 +2a(Xf-Xi), which is an equation I am familiar with.

    Using these formulas is easy. You simply list all things you know about an objects motion, then you can look at the equations and find what you can solve for. Often times you'll have to use 2-3 formulas to find what you are looking for.


    1. A car traveled at a constant speed and direction at 67.40 miles per hour for 3.00 hours, how far did the car travel.

    Using equation one, we see that the distance will equal the speed multiplied by the time, in this case 3 hours times 67.4miles per hour(notice the hours cancel and you get miles as your remaining units) which would equal 202 miles.

    2. A object accelerates from 0 to 60 meters per second in 5 seconds, what was its acceleration?

    Using equation two, we can see that acceleration equals change in velocity(0 to 60, so 60 meters per second) divided by change in time(0 to 5, so 5 seconds) note the units, it will be in meters per second squared. 60/5 is 12. So the answer would be 12 m/s^2.

    3. I drop a rock off a cliff, it takes 5 seconds to hit the ground, how high was the cliff.

    Using equation 3, we can see that the distance is going to be equal to its initial position(in this case we choose the top of the cliff as zero) + its initial velocity( I dropped the rock, so it has no Initial velocity) + 1/2acceleration(the earth's gravity would accelerate the object about 9.8 m/s^2) * time^2(5^2=25). Note that for any physics problem, you want to set up a coordinate plane. Which is exactly the same thing as the classic XY graph you've set up a million times in math before(assuming you have had much math) In this case I chose my x axis to be level with the top of the cliff and the positive y axis to be pointing DOWN, thus the direction the rock is falling, and gravity as well are both in positive directions, this is important.

    So, we get the height of the cliff to be height = 0 + 0 + 1/2(9.8m/s^2)(25 s^2). The second units cancel thus we get the height to be about 122 meters. I say "about" because there are a few things which would throw this value off a bit, such as wind resistance(which in this case we ignore), the centripetal force of the rotation of earth, and the fact the gravity will change the closer you get to earth(at this distance, it is a very small amount however).

    Its all actually very easy, its just plugging in numbers. As soon as you get to just the equations, you have the problem mostly solved, its setting up the problem and deciding which equation to use, or finding your own is normally the hard part.
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