A rocket ascends from rest in a uniform gravitational field by ejecting exhaust with constant speed u.
Assume that the rate at which mass is expelled is given by dm/dt=mk, where m is the instantaneous mass of the rocket and k is a cosntant, and that the rocket is ed by air resistance with a force bv where b is a constant. find the velocity of the rocket as a function of time.
I have worked it down to the last step but am having trouble finishing the integration to find v(t).
F_net = F_thrust - F_grav - F_air.resist
ma = (dp/dt) - mg - bv
(dv/dt) = (uk - g) - (bv/m)
m = m_o*e^kt ---> not sure if this part is right??
so (dv/dt) = (uk - g) - b*m_o*e^-kt*v
v = dx/dt so multiply by dt to get the integration equation of:
dv = (uk-g)dt - b*m_o*e^-kt*dx
my attempt so far:
v(t) = (uk-g)t - ????
***I am stuck on integrating the:
b*m_o*e^-kt*dx
I think that the answer comes out to be
v(t)=(uk-g)t - b*m_o*(e^-kt)*x(t)
because b and m_o are constants and i believe that e^-kt is taken as a constand as well. Leaving dx to be integrated between 0 and x(t).
am i correct? more opinions welcomed please
To integrate the expression b*m_o*e^(-kt)*dx, we need to use the integration by substitution method. Let's go through the steps:
First, we need to find dx in terms of dt. We know that v = dx/dt, so we rearrange the equation to get dx = v*dt.
The integral then becomes:
∫ b*m_o*e^(-kt)*v*dt.
Next, let's substitute u = -kt, and find du/dt:
du/dt = -k.
Rearranging the equation, we get dt = -du/k.
Substituting dt = -du/k into the integral, we have:
∫ b*m_o*e^u*v*(-du/k).
Simplifying, we obtain:
-b*m_o*v/k * ∫ e^u du.
Integrating ∫ e^u du gives us e^u. Hence, the integral becomes:
-b*m_o*v/k * e^u.
Substituting back u = -kt, we have:
-b*m_o*v/k * e^(-kt).
Finally, multiplying the original expression (uk - g)dt by t and integrating, we get:
v(t) = (uk - g)t - (b*m_o*v/k) * e^(-kt) + C,
where C is the constant of integration.
Therefore, the velocity of the rocket as a function of time is given by the expression:
v(t) = (uk - g)t - (b*m_o*v/k) * e^(-kt) + C.
To integrate the expression b * m_o * e^(-kt) * dx, you can use the method of substitution. Let's go through the steps:
1. First, let's rewrite the expression as b * m_o * e^(-kt) * dx = b * m_o * dx * e^(-kt).
2. To perform the integration, we need to find an appropriate substitution. Let's substitute u = -kt.
3. Taking the derivative of u with respect to x, we get du/dx = -k.
4. Rearrange the equation to isolate dx: dx = (-1/k) * du.
5. Now we can substitute these values into the integral:
∫ b * m_o * e^(-kt) * dx = ∫ b * m_o * dx * e^(-kt) = ∫ b * m_o * (-1/k) * du * e^u.
6. Simplify the expression inside the integral:
∫ b * m_o * (-1/k) * du * e^u = (-b * m_o / k) ∫ e^u du.
7. Now, we can integrate the term ∫ e^u du, which gives us the antiderivative e^u. Therefore:
∫ e^u du = e^u.
8. Plugging in the antiderivative, we have:
(-b * m_o / k) * e^u = (-b * m_o / k) * e^(-kt).
9. Finally, substitute back u = -kt:
(-b * m_o / k) * e^(-kt) = (-b * m_o / k) * e^(-kt).
Therefore, integrating b * m_o * e^(-kt) * dx yields (-b * m_o / k) * e^(-kt).