A differential equation for the velocity v of a falling mass...

A differential equation for the velocity v of a falling mass m subjected to air resistance proportional to the square of the instantaneous velocity is m dv/dt = mg - kv^2, where k > 0 is a constant of proportionality. The positive direction is downward. (a) Solve the equation subject to the initial condition v(0) = v_0. v(t) = sqrt(mg/k) tanh(sqrt(kg/m) t + tanh^-1((sqrt(k)v_0)/sqrt(mg))) (b) Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass. lim_{t->inf} v(t) = sqrt(mg/k) (c) If the distance s, measured from the point where the mass was released above ground, is related to velocity v by ds/dt = v(t), find an explicit expression for s(t) if s(0) = 0. s(t) = m/k ln cosh(sqrt(kg/k) t + tanh^-1((sqrt(k)v_0)/sqrt(mg))) - m/k ln cosh(tanh^-1((sqrt(k)v_0)/sqrt(mg)))

Transcript

00:01 So for this problem, we are told that a differential equation for the velocity of a falling mass m subjected to erresistant is proportional to the square of the instant stannius velocity.

00:14 So the equation that we are given is that the mass m times the derivative of the speed with respect to time is equal to the mass times the acceleration due to gravity minus the constant k, which is a constant of proportionality, times the speed square.

00:40 So the positive direction is downward.

00:44 So for part a of this problem, we need to solve the equation that is subjected to the initial condition b sub zero.

00:53 We are given that initial condition.

00:55 That b, the velocity at 0 at the time equals to 0 is equal to b sub 0.

01:03 So for this, we just write this equation down.

01:09 What we can do is to divide, we can multiply in here by minus.

01:16 So we are going to have the following and divide this by the constant, constant of proportionality divided by the mass, we put a minus in here, but also insert a minus in this product.

01:34 So we have this in here.

01:40 And now we just separate variables.

01:46 So what we are going to do is to pass this to the other side to divide the differential in the speed because this term in here depends on the velocity.

02:01 And this we pass the differential in time to the other side.

02:10 And now we take the integral in both sides of this equation.

02:17 So for the left side, we have the neparian logarithm of the velocity minus the mass times acceleration due to gravity, divided by the constant of proportionality, k.

02:31 And this is equal to minus k times the time t divided by the mass m, and this plus a constant of integration.

02:46 And now we just simply apply the exponential function to both sides of these equations.

02:51 So we are taking the velocity minus the mass times the acceleration due to gravity, divided by the constant of proportionality.

03:00 And this is equal to the exponential of minus the constant of proportionality k times the time t divided by the mass m times the exponential of c.

03:17 So this constant in here, the exponential of c, we're going to call this constant as simply b.

03:25 And then we can solve for the speed in terms of the time t so that we obtain that this is b times the exponential of minus the constant of proportionality k times the time t divided by the mass m and this plus the mass m times the acceleration due to gravity divided by the concept of proportionality k so in here in order to find the value of b we just simply evaluate the initial condition at zero at the time equals to zero that we know is b sub zero so if we we set in here, if we put in here the time equals to zero, we will obtain the exponential of zero that we know is one.

04:12 So from this, we're going to obtain that b -0 is equal to b plus m times acceleration due to gravity divided by the constant of proportionality k.

04:25 So from this, we obtain that b is equal to b -0 minus m times acceleration due to gravity, divided by the constant of proportionality k.

04:36 And then we substitute this into the expression, and finally we obtained that the speed is equal to b -0 minus n times the acceleration due to gravity divided by the constant of proportionality k times the exponential of minus k times the time t divided by the mass, and this plus the mass times the acceleration to to gravity divided by the constant k.

05:07 So that's a solution for part a of this problem.

05:13 And now for part c, we are asked to use the solution that we found in part a to determine the limiting or terminal velocity of the mass...

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