What Is Instantaneous Velocity?

[Miike012]

The question was, Explain instantaneous velocity.

My answer:
Limit as delta(t) approaches zero of average velocity is inst. velocity. Or as delta(t) approaches zero, average velocity approaches inst. velocity.
inst. velocity is the slope of the tangent line at a given point...

Then I wrote:
V = Lim Delta(Average Velocity)/Delta(t)
Delta(t) --> 0


Is my answer totally wrong because I got 5/10 pts on the answer?

 

[NascentOxygen]

tangent (of what graph) gives magnitude. velocity also needs a direction.

 

[Miike012]

This is the picture I added...

 

[Miike012]

Isnt saying slope of a tangent line at a "given point" general? It can be any point...

 

[Hootenanny]

The question was, Explain instantaneous velocity.

My answer:
Limit as delta(t) approaches zero of average velocity is inst. velocity. Or as delta(t) approaches zero, average velocity approaches inst. velocity.
inst. velocity is the slope of the tangent line at a given point...

Then I wrote:
V = Lim Delta(Average Velocity)/Delta(t)
Delta(t) --> 0Is my answer totally wrong because I got 5/10 pts on the answer?

Your answer isn't totally wrong - its only half wrong, that's why you got half marks .

The first thing to point out is that the average velocity is

$$v_\text{ave} = \frac{\Delta x}{\Delta t},$$

so your limit becomes

$$V = \lim_{\Delta t\to0} \left(\frac{\Delta x}{\Delta t}\right)/\Delta t = \lim_{\Delta t\to0} \Delta x,$$

which obviously isn't correct. So the limit you wrote, wasn't correct. The first part however, was okay:

Limit as delta(t) approaches zero of average velocity is inst. velocity. Or as delta(t) approaches zero, average velocity approaches inst. velocity.
inst.

However, your next sentence wasn't very clear:

velocity is the slope of the tangent line at a given point...

The tangent line to what? What are you plotting?

A better answer would have been as follows. The instantaneous velocity is equal to the average velocity in the limit as $\Delta t\to0$, or more generally, the derivative of position with respect to time. The velocity at a given point in time is equal to the gradient of the displacement at that point in time on a displacement-time graph.

In one dimension,

$$v = \lim_{t\to t_0} \frac{x(t_0)-x(t)}{t-t_0} = \frac{\text{d}x}{\text{d}t}.$$

Does that make sense?

 

[NascentOxygen]

This is the picture I added...

Well, there's your mistake. That slope is acceleration!

An "average velocity" graph doesn't make much sense (in this context). A velocity graph is velocity versus time, and that's instantaneous velocity, anyway. Or speed.

You were thinking of a position versus time graph, and finding the tangent to it.