Air resistance acting on a falling body can be taken into account by the approximate relation for the acceleration: a= dv/dt = g - kv, where k is a constant. (a) Derive a formula for the velocity of the body as a function of time assuming it starts from rest (v=0 at t=0). [Hint: Change variables by setting u=g-kv.] (b) Determine an expression for the terminal velocity, which is the maximum value the velocity reaches.
The answer has a question that not aware of the acceleration has an initial value.
To derive the formula for the velocity of the body as a function of time, we will follow the given hint and change variables by setting u = g - kv.
(a) Differentiating both sides of the equation, we get:
du/dt = d(g - kv)/dt
du/dt = -k * dv/dt
Since a = dv/dt, we can rewrite the equation as:
du/dt = -k * a
Now, substituting u = g - kv, we have:
du/dt = -k * a
du/dt = -k * (g - kv)
du/dt = -kg + k^2v
Rearranging the equation, we get:
du = -kg*dt + k^2v*dt
Integrating both sides of the equation, we obtain:
∫du = -k∫g*dt + k^2∫v*dt
u = -kgt + k^2∫v*dt
Since ∫v*dt is just the integral of velocity with respect to time, it gives us the displacement of the body.
Therefore, u = -kgt + k^2x
Now, let's solve for v:
u = g - kv
-gt + k^2x = g - kv
-gt = g - gk + k^2x
-gt = g(1 - k) + k^2x
Simplifying further:
-gt = g - gk + k^2x
-gt = g(1 - k + k^2x/g)
Dividing both sides by -g:
t = -1 + k - k^2x/g
Simplifying further:
t = -1 + k(1 - kx/g)
Rearranging the equation, we have:
t + 1 = k(1 - kx/g)
Dividing both sides by k:
(t + 1)/k = 1 - kx/g
Now, solving for x:
x = g/k - (t + 1)/k
Substituting this value of x into the equation for u, we get:
u = -kgt + k^2(g/k - (t + 1)/k)
Simplifying:
u = -kgt + k(g - (t + 1))
u = -kgt + kg - kt - k
u = -kgt - kt + kg - k
u = k(g - t(g + 1))
Now, substituting u = g - kv:
g - kv = k(g - t(g + 1))
Simplifying:
g - kv = kg - k^2t(g + 1)
Rearranging the equation:
kg - g = k^2t(g + 1) - kv
kg - g = k(kgt + k)(g + 1) - kv
Distributing:
kg - g = k^2g^2t + k^2g + kg + k - kv
Rearranging again:
kg - g = k^2g^2t + (kg - kv) + k
Since kg - g = k(g - 1):
k(g - 1) = k^2g^2t + k(g - 1) + k
Dividing both sides by k(g - 1):
1 = kg^2t + 1 + 1/g
Subtracting 1 from both sides:
0 = kg^2t + 1/g
Rearranging:
kg^2t = -1/g
Dividing both sides by kg^2:
t = -1/(kg^2)
This is the formula for the time as a function of the body's velocity.
(b) The terminal velocity is the maximum value the velocity reaches when the body falls. At terminal velocity, the acceleration is zero (a = 0).
So, we can set a = 0 in the original equation:
0 = g - kv
Solving for v, we have:
kv = g
v = g/k
Therefore, the expression for the terminal velocity is v = g/k.
To derive the formula for the velocity of the body as a function of time, we'll start by rearranging the given relation:
a = dv/dt = g - kv
Let's change variables by setting u = g - kv. Solving for v, we have:
v = (g - u) / k (Equation 1)
Now, let's differentiate u with respect to time:
du/dt = d(g - kv)/dt
Using the chain rule, we get:
du/dt = - k * dv/dt
Substituting a = dv/dt into this equation, we have:
du/dt = -k * a
Since u = g - kv, we can write du/dt as:
du/dt = d(g - kv)/dt = -kv' = -k * dv/dt
Substituting this into the above equation, we get:
-k * dv/dt = -k * a
Dividing both sides by -k, we have:
dv/dt = a
Now, let's substitute a with the given relation:
dv/dt = g - kv
Rearranging this equation, we have:
dv = (g - kv) * dt
Integrating both sides, we get:
∫dv = ∫(g - kv) * dt
Integrating the right side with respect to t, we have:
v = gt - kvt
Factoring out v from the right side, we get:
v = (g - kv)t
Substituting u = g - kv (from Equation 1), we have:
v = ut
This is the expression for the velocity of the body as a function of time.
Moving on to part (b), the terminal velocity is the maximum value that the velocity reaches. In other words, it is the value of v when acceleration is zero (a=0). Using the given relation:
a = g - kv
Setting a=0, we have:
0 = g - kv
Solving for v, we get:
kv = g
v = g / k
Therefore, the expression for the terminal velocity is v = g / k.
using their hint, let u = g-kv
du/dt = -k dv/dt
-1/k du/dt = u
du/u = -k dt
log u = -kt
u = e^(-kt)
g-kv = e^(-kt)
v = (g-e^(-kt))/k
= g/k - 1/k e^(-kt)
So, max v is g/k as t->infinity