# Differential Equations with a body initially at rest

1. Oct 16, 2011

### 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.

2. Oct 16, 2011

### 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.

3. Oct 16, 2011

### 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?

4. Oct 17, 2011

### dextercioby

Yes, it's ok. Excellent.