How can we determine velocity from a position-time graph without using calculus?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
oldspice1212
Messages
149
Reaction score
2
Hey guys, so I have a quick question about position - time graphs, so without using calculus for let's just say a particle moving west along a horizontal straight line every 0.10 s and the displacement is 0.022, 0.032, 0.042, etc.

Well the displacement from t = 0 aren't really good numbers that I used lol, but let's just assume it's non - linear, and without calculus would we just use a tangent line and find the slope of the tangent line to figure out velocity at what ever time interval we are given.


Thanks
 
Physics news on Phys.org
I think that is correct, the slope at that particular point is your velocity.
 
urbano said:
I think that is correct, the slope at that particular point is your velocity.

That's what I'm thinking but I wasn't certain.
 
Be careful. There are two different measures of velocity, and either one might be asked for, depending on the problem.

In both cases, assume that you already have a displacement vs. time curve.

  • The tangent of any point on the curve gives you the instantaneous velocity. That's the velocity of the particle at a particular instant in time.
  • But if you're given a particular time interval (meaning two, separate points on the curve), it usually means you are being asked to find the average velocity. For that, use

    [tex]\vec {v_{\mathrm{ave}}} = \frac{ \vec {\Delta s}}{\Delta t}[/tex]
    where [itex]\vec {\Delta s}[/itex] is the change in displacement and [itex]\Delta t[/itex] is the change in time (i.e., the specified time interval).

[Edit: Instantaneous velocity and average velocity become equal when [itex]\Delta t \rightarrow 0[/itex], at time t, where the instantaneous velocity was measured, meaning the two points on the curve merge into each other to form a single point. Otherwise, instantaneous and average velocities are not necessarily equal.]
 
Last edited: