Formulas for constant power acceleration

[rcgldr]

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]

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.

 

[rcgldr]

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]

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.

 

[rcgldr]

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