A skydiver has a downward velocity of v(t) = v_T(1 - exp(-gt/v_T)), where v_T is

Question

A skydiver has a downward velocity of v(t) = v_T(1 - exp(-gt/v_T)), where v_T is the terminal velocity of the skydiver and g is the acceleration due to gravity. If the skydiver has a terminal velocity of 50.0 m/s, how far will she fall between t = 0 and t = 3.0 s? What will be her average velocity during this time interval? Consider an object which has a position x_0 and a velocity v_0 at t = 0. Show that if the object has a constant acceleration a, it will have a velocity v(t) = v_0 +at and a position x(t) = x_0 + v_0t + 1/2 at^2.

Explanation / Answer

When a body falls, it is acted upon by the gravitational force acting downwards and the force due to air resistance upwards.

Fair  must be a function of velocity because it is a frictional force ,so we take it as -kv.(k will have the dimension of [MT-1]

So total force is,

F = Fg + Fair

--> ma = mg - kv

--> dv/dt = g - (k/m)v {after dividing the whole equation by m}

--> dv/dt = (k/m)[(mg/k)-v]

The above differential equation can be solved by taking the bracketed term as a new v' and dv'/dt = dv/dt

--> dv/dt = (k/m)v' or dv/v' = (k/m)dt

If you integrate the above equation and solve it you get,

v' = cexp(-kt/m) --> v = mg/k + cexp(-kt/m)

After finding c by applying the initial conditions,

v = mg/k(1- exp(-kt/m))

As t goes to infinity, v = mg/k which is the terminal velocity vt

v = vt(1-exp(-gt/vt))

Terminal velocity is the velocity attained by the body as time tends to infinity.

So now you understand the formula, now to the QN...

3. We have to find the distance traveled by the diver as she falls through the air. If the velocity was constant, then distance traveled would be distance = velocity*time taken.

Unfortunately it isn't constant. What we can do however is to take smaller time intervals and assume that the distance is constant in that time interval. Let's say we take t =0 to t=0.5,t=0.5 to t=1 and so on and we find the velocity in each interval. If we plot a velocity-time graph, then the area under that graph would give you the total distance traveled. Accuracy increases as the interval is made smaller. This is the essence of integration!

Now doing all of that would take a long time, instead we do something simpler- we integrate the velocity function within the given time limits.

dx/dt = vt(1-exp(-gt/vt)

Integrating the above expression,

x= int{vt(1-exp(-gt/vt)}

= vt[t|03 + (vt/g)exp(-gt/vt)|03]

Applying the limits and substituting the given values,

x = 50* [3+(50/9.8)*{exp(-9.8*3/50) - 1} ]

= 50 * [3 + (5.1)*{0.555-1}]

= 50 * [3 - 2.2695]

= 36.525m

This will be the distance traveled.

Avg velocity = (v(3)-v(0))/(3-0) = 22 m/s

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