Differential Equations with a body initially at rest

[e to the i pi]

1. A body of mass 2kg is initially at rest and is acted upon by a force of (v - 4) Newtons where v is the velocity in m/s. The body moves in a straight line as a result of the force.



2. a. Show that the acceleration of the body is given by dv/dt = (v - 4) / 2
b. Solve the differential equation in part a to find v as a function of t.



3. a. I used the formula F = ma where F = (v - 4) and m = 2
(v - 4) = 2a
a = (v - 4) / 2

b. I tried to solve it like any other differential equation with the following initial conditions:
when t = 0, v = 0
But I found it very difficult and challenging:
dv/dt = (v - 4) / 2
2 dv/dt = v - 4
2 / dt = (v - 4) / dv
I want to change the division sign to a multiplication sign so that I can take the integral of both sides, but I don't know how to algebraically manipulate it to be in that form.

 

[dextercioby]

You don't really know how to separate variables.

$$\frac{dv}{dt} = \frac{v-4}{2} \Rightarrow \frac{dv}{v-4} = \frac{1}{2}dt$$

Now integrate and use the initial condition.

 

[e to the i pi]

So after that I do this:
∫1 / (v - 4) dv = ∫1/2 dt
ln|v - 4| = 1/2 t + C
when t = 0, v = 0
ln|-4| = C
C = ln(4)
So this proves that we need the negative solution of the absolute value:
ln(4 - v) = 1/2 t + ln(4)
ln(4 - v) - ln(4) = 1/2 t
ln((4 - v) / 4) = 1/2 t
Since it has a base of e and a power of (1/2 t):
e^(1/2 t) = (4 - v) / 4
4e^(1/2 t) = 4 - v
v = 4 - 4e^(1/2 t)
Is that all correct?

 

[dextercioby]

Yes, it's ok. Excellent.

 

[asteriatic]

I can confirm that the approach taken in part 3a is correct. The formula F = ma is the basic equation for Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In this case, the net force is (v-4) and the mass is 2kg, resulting in an acceleration of (v-4)/2.

In order to solve the differential equation in part 3b, we can use the method of separation of variables. This involves separating the variables v and t on opposite sides of the equation and then integrating both sides. So, we have:

2 dv / (v-4) = dt

Integrating both sides, we get:

2 ln|v-4| = t + C

where C is the constant of integration. We can rewrite this as:

ln|v-4| = t/2 + C/2

Using the initial condition v(0) = 0, we can solve for C and get C = -ln4. Substituting this back into the equation, we get:

ln|v-4| = t/2 - ln4
v-4 = e^(t/2 - ln4)
v-4 = e^(t/2) / 4
v = e^(t/2) / 4 + 4

Therefore, the velocity of the body as a function of time is v(t) = e^(t/2) / 4 + 4.

In conclusion, the differential equation in part 3a can be solved using the method of separation of variables to find the velocity of the body as a function of time. This approach is commonly used in physics and engineering to solve problems involving motion and forces.