Formulas for constant power acceleration

Homework Helper
This isn't homework. A friend asked about this so I decided to work out the formulas, but wanted to know if this was already done by someone here (otherwise I'll do the math).

p = power (constant)
a = acceleration
v = velocity
x = position
t = time
f = force

Assume an object is initially at rest, at position zero and time zero:

v0 = 0
x0 = 0
t0 = 0

f = m a
p = f v
f = p / v

first step

a = f / m = dv/dt = p / (m v)
v dv = (p/m) dt
1/2 v2 = (p/m) t

$$v = \frac{dx}{dt} = \sqrt {\frac{2\ p\ t}{m}}$$

This is continued to find x as a function of t, then t as a function of x

Then determine f(x) = p / v(x)

and finally show that work done is

$$p\ t_1 = \int_0^{x_1} f(x) dx$$

mfb
Mentor
You can simply integrate your equation for v(t) and get x(t). That can be used to get v(x) and the other expressions.

Homework Helper
You can simply integrate your equation for v(t) and get x(t). That can be used to get v(x) and the other expressions.
I know that, was just wondering if someone here had already done this in a previous thread. The previous threads I did find never actually completed the formulas. I'll go ahead and do this later.

mfb
Mentor
I am sure this has been done before, it is a nice and easy problem in mechanics and can be solved with very basic concepts.

Homework Helper
$$p = f \ v$$
$$a = \frac{dv}{dt} = \frac{f}{m} = \frac {p} {m\ v}$$
$$v\ dv = \frac{p}{m}\ dt$$
$$\frac{1}{2}\ v^2 = \frac{p}{m}\ t$$
$$v = \frac{dx}{dt} = \sqrt {\frac{2\ p\ t}{m}}$$
$$dx = \sqrt {\frac{2\ p\ t}{m}}\ dt$$
$$x = \sqrt {\frac{8\ p\ t^3}{9\ m}}$$
$$t = \sqrt {\frac{9\ m\ x^2}{8\ p}}$$
a as a function of t:
$$a = \frac {p} {m\ v} = \frac {p} {m\ {\sqrt {\frac{2\ p\ t}{m}}}} = \sqrt {\frac{p}{2\ m\ t}}$$
v as function of x:
$$v = \sqrt {\frac{2\ p\ \sqrt {\frac{9\ m\ x^2}{8\ p}}}{m}} = \sqrt {\frac{3\ p\ x}{m}}$$
a as a function of x:
$$a = \frac{p}{m \ v} = \frac{p}{m \ \sqrt {\frac{3\ p\ x}{m}}} = \sqrt {\frac{p^2}{3\ m^2\ x}}$$
f as a function of x:
$$f = m\ a = \sqrt {\frac{m\ p^2}{3\ x}}$$
work done versus x:
$$w = \int_0^x \sqrt {\frac{m\ p^2}{3\ x}} \ dx = \sqrt{\frac{9\ m\ p^2\ x^2}{8}}$$
work done versus time:
$$w = \sqrt{\frac{9\ m\ p^2\ \left( \sqrt {\frac{8\ p\ t^3}{9\ m}} \right )^2}{8}} = p \ t$$

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