We would like to know the velocity of the block when it reaches some position x. Finding this requires an integration. However, acceleration is defined as a derivative with respect to time, which leads to integrals with respect to time, but the force is given as a function of position. To get around this, use the chain rule to find an alternative definition for the acceleration ax that can be written in terms of vx and dx/ dx.

Answer:

An alternative definition for the acceleration ax that can be written in terms of [tex]v_x[/tex] and [tex]\frac{dv_x}{dx}[/tex] is [tex]a_x=v_x \frac{dv_x}{dx}[/tex]

Step-by-step explanation:

We know that :

[tex]a_x=\frac{dv_x}{dt}[/tex]

Now we are supposed to find an alternative definition for the acceleration ax that can be written in terms of [tex]v_x[/tex] and [tex]\frac{dv_x}{dx}[/tex]

So, We will use chain rule over here :

[tex]a_x=\frac{dv_x}{dt}\\a_x=\frac{dv_x}{dt} \times \frac{dx}{dx}\\a_x=\frac{dv_x}{dx} \times \frac{dx}{dt}\\a_x=\frac{dv_x}{dx} \times \frac{dx}{dt} [\frac{dx}{dt}=v_x]\\a_x=\frac{dv_x}{dx} \times v_x\\a_x=v_x\frac{dv_x}{dx}[/tex]

Hence an alternative definition for the acceleration ax that can be written in terms of [tex]v_x[/tex] and [tex]\frac{dv_x}{dx}[/tex] is [tex]a_x=v_x \frac{dv_x}{dx}[/tex]

The alternative definition for the acceleration ax that can be written in terms of vx and dx/ dx is [tex]a_x = v_x \frac{dv_x}{dx}[/tex]

Here we used the chain rule:

It is the basic method for differentiating a composite function. Here the first factor on the right, Df(g(x)), represents that the derivative of f(x) is first found and then x, wherever it arises, is replaced by the function g(x).

So,

[tex]a_x = \frac{dv_x}{dt}\\\\ = \frac{dv_x}{dt} \times \frac{dx}{dx}\\\\= \frac{dv_x}{dx} \times \frac{dx}{dt}\\\\ = \frac{dv_x}{dx} \times \frac{dx}{dt} (\frac{dx}{dt} = v_x)\\\\ = \frac{dv_x}{dx} \times v_x\\\\= v_x\frac{dv_x}{dx}[/tex]

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