Instantaneous velocity is slope of displacement-time curve?

I have a toss-up question that I'm not sure how to answer. Here it is (true or false):

The instantaneous velocity at a particular time is the slope of the displacement-time curve at that position.

I know that this statement would be true if it said position-time curve because that would mean that the change in position is the displacement. However, as the statement currently stands, it is saying that instantaneous velocity is equal to the change in the displacement over time, which is the change in the change in position over time. Am I correct to say that?

But when I do an example of each, they both seem to work. Like this:

displacement 1 at 1 second is 10 m from the starting position.

displacement 2 at 2 seconds is 20 m from the starting position.

The change in displacements is 10 m, and over 1 second. So the velocity over that time period is 10 m/s.

And the same with position:

position 1 at 1 second is 10 m from the starting position.

position 2 at 2 seconds is 20 m from the starting position.

The change in position is 10 m, and over 1 second. So the velocity over that time period is 10 m/s.

I can't decide what makes sense here. Is the statement true or false?

If your position is in 3D space (which means your position vector must be defined), then there is no distinction between displacement and change in position.

$s=\boldsymbol{R_f-R_i}=\Delta\boldsymbol{R}$ , where $s$ is displacement and $R$ is position.

However, in $v = ds/dt$, $ds$ does not mean change in displacement but rather an infinitesimally small displacement (infinitesimally small change in position). Since $v=ds/dt$, so the instantaneous velocity is the gradient to the tangent of the curve at that particular point.

In my opinion I believe it's just an issue of wording. I don't think change in displacement means $ΔΔR$, but rather the actual displacement(change in position) that occurred during a time $ΔT$

• It's in 1D (going in one direction). And I know what you said is true, but what about the change in displacement? Is that the same as the change in position? I know that displacement = change in position. But it just seems odd that change in displacement also = change in position. – Michael Yaworski Oct 10 '14 at 20:21
• In my opinion I believe it's just an issue of wording. I don't think change in displacement means $\Delta\Delta{R}$, but rather the actual displacement(change in position) that occurred during a time $\Delta{T}$ – t.c Oct 10 '14 at 20:25
• I agree. So if you picture a displacement-time curve, would you say that any tangent to that curve has slope that's instantaneous velocity? That's the real question and I think that's correct. – Michael Yaworski Oct 10 '14 at 20:31
• Yes, that is true, since $v = ds/dt$, so the instantaneous velocity is the gradient to the tangent of the curve at that particular point. – t.c Oct 10 '14 at 20:33
• Okay, thanks. Maybe you can incorporate these comments into your answer so that I can accept it. – Michael Yaworski Oct 10 '14 at 20:34