11Fvariableforces


 * Variable Forces**

Recall from chapter 8, the following rules to use when accleration is variable:

math \\ \text{If } a=f(t) \\. \\ . \qquad \text{Use } a=\dfrac{dv}{dt} \text{ or } a=\dfrac{d^2x}{dt^2} math

math \\ \text{If } a=f(x) \\. \\ . \qquad \text{Use } a=\dfrac{d}{dx} \Big( \frac{1}{2}v^2 \Big) math

math \\ \text{If } a=f(v) \\. \\ . \qquad \text{Use } a=\dfrac{dv}{dt} \; \; \textit{ if initial conditions are in terms of v and t} \\. \\ . \qquad \text{Use } a=v\dfrac{dv}{dx} \; \; \textit{ if initial conditions are in terms of v and x} math

**Example 1**

An object of mass 2kg is acted on by a force whose direction is constant and whose magnitude at time, t, is given by (6 – 2t) Newtons. The object has an initial velocity of +2m/s (in the same direction as the force). Find the velocity after 4 seconds.

__**Solution:**__


 * ma = R
 * 2a = 6 – 2t
 * a = 3 – t

{a is in terms of t so we can antidifferentiate immediately}

math \\ . \qquad \dfrac{dv}{dt} = 3 - t \\. \\ . \qquad v=\displaystyle{\int 3-t \; dt} \\. \\ . \qquad v=3t-\frac{1}{2}t^2+c math

At t = 0, ... ... we have that v = 2, ... ... so c = 2.

math \\ . \qquad v=3t-\frac{1}{2}t^2+2 \\. \\ . \qquad v(4)=3 \times 4 - \frac{1}{2}4^2+2 \\. \\ . \qquad v(4)=6 \;\; \text{m/s} math

**Example 2**

A resultant force of 3__**i**__ – 4__**j**__ acts on a body with a mass of 2 kg. Initially, the body has a velocity of 2__**i**__


 * a) Find the acceleration **

... ... m__a__ = __R__ ... ... 2__a__ = 3__i__ – 4__j__ ... ... __a__ = 1.5__i__ – 2__j__

... ... __a__ = 1.5__i__ – 2__j__ ... ... __v__ = 1.5t__i__ – 2t__j__ + __c__
 * b) Find the velocity **

... ... Initially we have __v__ = 2__i__ ... ... so therefore __c__ = 2__i__

... ... __v__ = (1.5t + 2)__i__ – 2t__j__

... ... __v__ = (1.5t + 2)__i__ – 2t__j__ ... ... __s__ = (0.75t 2 + 2t)__i__ – t 2 __j__ + __d__
 * c) Find the displacement from the original position **

... ... Initially we have displacement of 0, so __d__ = 0

... ... __s__ = (0.75t 2 + 2t)__i__ – t 2 __j__


 * d) Find the displacement after 2 seconds **

... ... __s__(2) = (0.75 x 2 2 + 2 x 2)__i__ – 2 2 __ j __

... ... __s__(2) = 7__i__ – 4__j__

**Example 3**

A resultant force of 4x + 5 acts on a body of mass 2kg moving in a straight line where x is the displacement from a fixed origin. If the velocity, v, is 3 when x is 0, find v when x is 1.

__**Solution:**__

Using ma = R gives us:

... ... 2a = 4x + 5 ... ... a = 2x + 2.5

Since a = f(x), from the list at the top of the page we should use:

math \\ . \qquad a=\dfrac{d}{dx} \Big( \frac{1}{2}v^2 \Big) \\. \\ . \qquad \dfrac{d}{dx} \Big( \frac{1}{2}v^2 \Big)=2x+2.5 \\. \\ . \qquad \frac{1}{2}v^2=\displaystyle {\int 2x+2.5 \; dx} \\. \\ . \qquad \frac{1}{2}v^2=x^2+2.5x+c math

math \\ . \qquad \text{When } x=0, \; v=3, \; \text{ so } \; c=4.5 \\. \\ . \qquad \frac{1}{2}v^2=x^2+2.5x+4.5 \\. \\ . \qquad v^2=2x^2+5x+9 math

This is the relation between v and x.

Now answer the question: find v when x = 1 math \\ . \qquad v^2=2 \times 1^2+5 \times 1+9 \\. \\ . \qquad v^2=16 \\. \\ . \qquad v=\pm 4 \;\; m/s math

{there are two answers to the question.}

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