Thread: Instantaneous Velocity!

How do you calculate it is? is it

Velocity inst. = (df - di)/(tf - ti)

or what??

and instantaneous velocity is the average velocity between two points right?

Originally Posted by sakonpure6

How do you calculate it is? is it

Velocity inst. = (df - di)/(tf - ti)

or what??

and instantaneous velocity is the average velocity between two points right?

Not quite.

Average velocity = (d_f - d_i)/(t_f - t_i)

Instantaneous velocity is the velocity at one point.
It is what you get if pick your t_i and t_f at the same time.
You can't really do that since you'd be dividing by zero, but you can make the one approach the other as close as you want.
This is what you do when you calculate a limit.

Originally Posted by ILikeSerena

Not quite.

Average velocity = (d_f - d_i)/(t_f - t_i)

Instantaneous velocity is the velocity at one point.
It is what you get if pick your t_i and t_f at the same time.
You can't really do that since you'd be dividing by zero, but you can make the one approach the other as close as you want.
This is what you do when you calculate a limit.

lets say i have those two points (1,20) and (2,40)

would that make the instantaneous velocity 20 at 1.5?

Originally Posted by sakonpure6

lets say i have those two points (1,20) and (2,40)

would that make the instantaneous velocity 20 at 1.5?

It would mean that you cannot find the instantaneous velocity without more information.

Now if for instance your acceleration is zero, than your velocity will be the same everywhere.
The instantaneous velocity would be the same as the average velocity then.

well i have a set of data that makes up an increasing exponential function and i am supposed to find the acceleration of it and to do that my teacher said to take the instantaneous velocity between all two data points which would give me a linear function of where i can determine the acceleration by finding the slope.

So i have the first two data points (0s , 0 cm [E]) and ( 0.05s, 0.80 cm [E]) , i determined that the instantaneous velocity between those points is 16 cm/s [E] at 0.025s. Is that right?

What sort of set of data are you talking about?

I think I understand.

You have a set of data with a number of consecutive times, probably separated by 0.05 seconds.
At each time you have a distance traveled in centimeters.

Indeed, in that case the approximation of the instantaneous velocity is the average velocity between 2 consecutive points.
This works because those 2 points are close enough together.

so my answer of 16 is right?

Originally Posted by sakonpure6

so my answer of 16 is right?

Yes.

oh great! thank you very much!!!!!!!!!!!!!!!!