8Eairresistance


 * Kinematics and Differential Equations**

Motion in a Resisting Medium

When an object travels through air (or falls), the air resists the motion and acts to slow it down. ie the medium retards the motion. It always acts opposite the direction of motion.

In general, air resistance is proportional to some power of the speed, kv n, where k and n are constants.

If an object falls far enough, it will gain enough velocity so that air resistance will cancel out the acceleration due to gravity and the object falls at a constant velocity. This is called **terminal velocity**.

Example

A body falls from rest with an acceleration given by math . \qquad a=g-0.2v \;\; m/s^2 math

Find an expression for the velocity at any time t.

From rest, so v = 0 at t = 0 math . \qquad \text{Given } a=f(v), \; \text{ use } \; a=\dfrac{dv}{dt} math

math \\ .\qquad \dfrac{dv}{dt}=g-0.2v \\. \\ . \qquad \dfrac{dt}{dv}=\dfrac{1}{g-0.2v} math

integrate math \\ . \qquad t=\displaystyle{\int} \dfrac{1}{g-0.2v} \,dv \\. \\ . \qquad t=-5\displaystyle{\int} \dfrac{-0.2}{g-0.2v} \,dv \\. \\ math

We now have a function in the denominator and its derivative as the numerator math . \qquad t=-5\log_e \big| g-0.2v \big| +c math

From the situation g – 0.2v must be positive, so don't need absolute value

When t = 0, v = 0 so c = 5log e g math \\ . \qquad t=-5\log_e(g-0.2v)+5log_eg \\. \\ . \qquad t=5\log_e \dfrac{g}{g-0.2v} \\. \\ . \qquad \dfrac{t}{5} = \log_e \dfrac{g}{g-0.2v} math

By log laws math \\ . \qquad e^{\frac{1}{5}t}=\dfrac{g}{g-0.2v} \\. \\ . \qquad (g-0.2v)e^{\frac{1}{5}t}=g \\ .\\ . \qquad ge^{\frac{1}{5}t} - 0.2ve^{\frac{1}{5}t} =g math math \\ . \qquad -0.2ve^{\frac{1}{5}t}=-ge^{\frac{1}{5}t}+g \\ .\\ . \qquad ve^{\frac{1}{5}t}=5ge^{\frac{1}{5}t}-5g math

math \\ . \qquad v=5g-5ge^{-\frac{1}{5}t} \\. \\ . \qquad v=5g \left( 1-e^{-\frac{1}{5}t} \right) math

Notice on the graph that **terminal velocity** appears as an asymptote.

In the rule for v, math \\ . \qquad \text{As } t \, \rightarrow \, \infty, \, e^{-\frac{1}{5}t} \, \rightarrow \, 0 \\. \\ . \qquad \text{so } v \, \rightarrow \, 5g math

So **terminal velocity** is 5g = 49 m/s

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