A differential equation governing the velocity v of a falling mass m
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 initial condition v(0)=v_0.
(b)Determine the limiting velocity.
(c)If the distance s, measured from the point where the mass was released above the ground, is related to velocity by ds/dt=v, find an explicit form of s if s(0)=0.
Still would appreciate someone pointing us in the right direction :/
no ideas on these problems? Homework dumps are discouraged.
To solve the given differential equation, we need to separate the variables and integrate both sides.
(a) Solve the equation subject to the initial condition v(0) = v₀:
The given differential equation is:
m(dv/dt) = mg - kv²
Rearranging the terms, we get:
dv/(mg - kv²) = dt/m
Now, we can separate the variables by multiplying by dt and dividing by (mg - kv²):
(1/(mg - kv²)) dv = dt/m
Integrating both sides with respect to their respective variables, we get:
∫(1/(mg - kv²)) dv = ∫(1/m) dt
To integrate the left side, we can use a partial fraction decomposition:
1/(mg - kv²) = A/(mg - kv) + B/(mg + kv)
Here, A and B are constants that we need to determine. We can find these constants by finding the common denominator and equating the numerators on both sides:
1 = A(mg + kv) + B(mg - kv)
Expanding and equating coefficients, we get two equations:
0 = A + B (coefficient of v)
1 = Amg - Bmg (coefficient of constant term)
From the first equation, we can see that A = -B. Substituting this into the second equation, we get A = -1/2 and B = 1/2.
Now, we can rewrite the integral as:
∫((-1/2)/(mg - kv) + (1/2)/(mg + kv)) dv = ∫(1/m) dt
Integrating both sides, we get:
(-1/2)ln|mg - kv| + (1/2)ln|mg + kv| = (1/m)t + C
Here, C is the constant of integration. We can simplify this equation as:
ln|mg + kv| - ln|mg - kv| = (2/m)t + 2C
Using the logarithmic property ln(a) - ln(b) = ln(a/b), we can further simplify:
ln|mg + kv|/(mg - kv) = (2/m)t + 2C
Now, we can exponentiate both sides:
|mg + kv|/(mg - kv) = e^(2t/m) * e^(2C)
Since e^(2C) is a constant value, we can replace it with a new constant A:
|mg + kv|/(mg - kv) = A * e^(2t/m)
Now, we consider the absolute value term. When mg + kv > 0, we can drop the absolute value:
(mg + kv)/(mg - kv) = A * e^(2t/m)
Similarly, when mg + kv < 0, we can remove the negative sign:
-(mg + kv)/(mg - kv) = A * e^(2t/m)
We can now solve these two cases separately:
Case 1: mg + kv > 0
(mg + kv)/(mg - kv) = A * e^(2t/m)
Cross-multiplying:
mg + kv = A * (mg - kv) * e^(2t/m)
Expanding and rearranging:
(kA)e^(2t/m) = (A - 1)mg - A(kv)
Dividing both sides by kA:
e^(2t/m) = [(A - 1)mg - A(kv)] / (kA)
e^(2t/m) = [(A - 1)mg/k - v]
Taking the natural logarithm of both sides:
2t/m = ln[(A - 1)mg/k - v]
Simplifying:
t = (m/2) * ln[(A - 1)mg/k - v] + C₁
Note that C₁ is a constant of integration and can be combined with the constant term A into a new constant B.
Case 2: mg + kv < 0
-(mg + kv)/(mg - kv) = A * e^(2t/m)
Cross-multiplying and following similar steps as above:
e^(2t/m) = [A(mg - kv) - (1 - A)mg] / (kA)
e^(2t/m) = [(A - 1)mg + kv]
2t/m = ln[(A - 1)mg + kv]
t = (m/2) * ln[(A - 1)mg + kv] + C₂
Here, C₂ is another constant of integration.
(b) Determining the limiting velocity:
To find the limiting velocity, we consider the behavior as t approaches infinity. As t approaches infinity, the exponential terms e^(2t/m) will become much larger. Therefore, the other terms must approach zero for the equation to remain balanced.
From the first case, we can observe that [(A - 1)mg - v] must approach zero as t approaches infinity. This means, [(A - 1)mg] must approach v, or we can rewrite it as:
(A - 1)mg ≈ v
Simplifying, we get:
(A - 1) ≈ v/(mg)
As v is the limiting velocity, we can substitute it into the equation:
(A - 1) ≈ v/(mg)
As t approaches infinity, the value of (A - 1) must approach zero to balance the equation. Therefore, we can ignore the constant term 1 in the expression (A - 1) and rewrite it as:
A ≈ v/(mg)
Now, substituting A into the equations from both cases:
t ≈ (m/2) * ln[(v/(mg) - 1)mg/k - v] + C₁ (Case 1)
t ≈ (m/2) * ln[(v/(mg) - 1)mg + kv] + C₂ (Case 2)
Since ln[(v/(mg) - 1)mg/k - v] and ln[(v/(mg) - 1)mg + kv] approach infinity as t approaches infinity, we can ignore the other terms in these equations. Therefore, the limiting velocity occurs when |v/(mg) - 1| approaches zero.
If v/(mg) ≈ 1, then the limiting velocity is approximately mg.
(c) If the distance s, measured from the point where the mass was released above the ground, is related to velocity by ds/dt = v, we can find an explicit form of s if s(0) = 0:
From the given relation, we have:
ds/dt = v
Integrating both sides with respect to their respective variables, we get:
∫ds = ∫v dt
The left side represents the distance traveled (s), and the right side represents the integral of velocity (v) with respect to time (t):
s = ∫v dt
Now, we need to find the integral of v with respect to t. We can substitute the value of v from the previous solutions:
For Case 1:
v = (mg - (A - 1)kv) * e^(-2t/m)
For Case 2:
v = (-(mg + kv) = (A(mg - kv)) * e^(-2t/m)
Substituting these expressions for v into the integral:
For Case 1:
s = ∫[(mg - (A - 1)kv) * e^(-2t/m)] dt
For Case 2:
s = ∫[(A(mg - kv)) * e^(-2t/m)] dt
Integrating both sides will give you the explicit form of s in each case.