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The question is a three part question. Part a) asked to find the average velocity with the equation $y= -\frac{1}{25}x^2 + \frac{4}{5}x$ on the interval $[6,10]$. Part b) asked to find the instantaneous velocity at $x=6$ and $x=10$. Part c) asks to find the acceleration at $x=4$. So I am not sure what to do after part b).

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Isn't this question not well posed? The interval of definition is [6,10] and the request is to find the acceleration at $x=4$. –  user44676 Oct 14 '12 at 22:11
    
It says to find the average velocity on that interval --- it doesn't say the function is defined only on that interval. –  Gerry Myerson Oct 15 '12 at 12:29

3 Answers 3

Velocity is the instantaneous rate of change of position. Assuming $y(t)$ is position-time, we have

$$V(x) = \frac{dy}{dx} = -\frac{2}{25}x+\frac{4}{5}$$

and acceleration is the instantaneous rate of change of velocity:

$$A(x) = \frac{dV}{dx} = -\frac{2}{25}$$

Simply find now $A(4)$.

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Assuming that $y$ is meant to stand for a distance, and $x$ is meant to stand for time, then, as you probably know, instantaneous velocity is the rate of change of $y$ with respect to $x$. Acceleration is the rate of change of velocity with respect to $x$. So if you know what to do to $y$ to get the velocity, then you know what to do to the velocity to get the acceleration.

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Acceleration is another way of saying rate of change of velocity--while velocity is rate of change of position--so acceleration is the second derivative. Hence, the answer will be $$\frac{d^2y}{dx^2}|_{x=4}.$$

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