Can someone explain instantaneous velocity please
Favorites|Homepage
Subscriptions | sitemap
HOME > Mathematics > Can someone explain instantaneous velocity please

Can someone explain instantaneous velocity please

[From: ] [author: ] [Date: 11-07-01] [Hit: ]
I know that d is for derivative, but I have not studied derivatives yet.The textbook also noted that for any n, the derivative of t^n is nt^(n-1). I suppose this means that the derivative of t^2 is 2t? but wouldnt the 2ts cross out if placed within the equation?......
So the equation given is x=20+5t^2
x= distance and t=time. I am supposed to find the instantaneous velocity using the equaiton v=dx/dt
I know that d is for derivative, but I have not studied derivatives yet.
The textbook also noted that for any n, the derivative of t^n is nt^(n-1). I suppose this means that the derivative of t^2 is 2t? but wouldn't the 2t's cross out if placed within the equation? Is d in dx different from d in dt? I am obviously missing something. Could someone please explain this to a student who has not begun precalc?
Thank you for your time.

-
Yes, what you're missing is that the "d" in dx/dt is not a variable. You have to read the whole thing as one unit. Or you can read "d/dt" as one thing, an operation called "differentiate with respect to time" and it's operating on x.

It's a notation for a particular kind of operation. And when you apply the "d/dt" operation on the function (20 + 5t^2) it produces the function 10t.

That's because it produces 5*df/dt from 5f, and it produces 0 from the constant 20 and 2t from t^2. So d/dt[20 + 5t^2] = 0 + 5*(2t) = 10t.

-
The instantaneous velocity for this equation is the first derivative.

v = 10t

now for any time you can find the instantaneous by plugging in for t.

-
No you are substituting 2t for t²
You also need to know that the derivative of a constant is zero

so dx/dt = 5(2t) = 10t

This is a poor question if you don't have calculus.

-
v = dx/dt = 10t

This gives the velocity at any specific time.

Reading up on 'differentiating polynomials' will help further.

:)>

-
You can think of dx/dt as a very small ratio of delta x and delta t -> this means that for a very small amount of change in t (this value being dt), a very small change in x occurs (this value is dx).

Using the familiar equation speed = distance / time, you can see that the instantaneous velocity is equal to the very small dx divided by the very small dt:

v = dx / dt

As for the derivative itself, you have the rule correct. If x = 20 + 5t^2, then dx / dt = 10*t

The dt in the left side of the equation is not a true value, which is why you cannot put it on the right side of the equation. dx/dt is simply a way of denoting "the derivative of x with respect to t".
1
keywords: velocity,please,explain,Can,instantaneous,someone,Can someone explain instantaneous velocity please
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .