Displacement from variable forces

Question

Displacement from a singular force over time is given by the equation $${1\over2}{F\over m}t^2 $$ Where F is force, m is mass, and t is time.

But what if F is variable over time?

My best guess is to find the "area" under the curve of F, as on a graph, using integration by the trapezoidal method, with respect to t

This gives SI units of joules, so then we must divide by F to get total displacement over time.

Is this the correct solution?

Answer 1

No. The correct statement (Newton's second law) is that $$F=\frac{dp}{dt}$$ where $p$ is the momentum of the object. In the non-relativistic case, and with non-varying mass, this simplifies to $$F=ma,$$ with $a$ the acceleration, the second derivative of displacement with respect to time. In general this differential equation may be hard to solve.

Answer 2

$F(t)=m\frac{d^2x}{dt^2}$

For simplicity, Let, $F(t)=at+k$ where, a,k are constants

$\int F(t)dt=m\int (\frac{d^2x}{dt^2})dt$

Now if i need to calculate the distance from t=0 to t =2 sec

$\int F(t)dt=m\int (\frac{d^2x}{dt^2})dt$

$\int (at+k)dt=m\int (\frac{d^2x}{dt^2})dt$

$a\frac{t^2}{2}+k{t}+b =m(\frac{dx}{dt})+c$

Now if i need to calculate the distance from t=0 to t =2 sec

$\int_0^2(a\frac{t^2}{2}+k{t}+b-c)dt =m\int_{x_1}^{x_2}dx$

$[a\frac{t^3}{6}+k\frac{t^2}{2}+(b-c)t]^2_0 =m(x_2-x_1)$

$X=(x_2-x_1)=\frac{a\frac{2^3}{6}+k\frac{2^2}{2}+(b-c)2}{m}$

Answer 3

Newton's Second Law says that $m\frac{d^2x}{dt^2} = F$. Let $t_0$ be some time, and $x_0, v_0$ the inital position and velocity. Suppose that $F$ changes with time, that is, $F = F(t)$. Then:

$$a(t) = \frac1mF(t)$$

$$v(t) - v(t_0) = \int_{t_0}^ta(t)\ dt = \frac1m \int_{t_0}^t F(t)\ dt$$

$$v(t) = v_0 + \frac1m\int_{t_0}^t F(t)\ dt$$

$$x(t) - x(t_0) = \int_{t_0}^tv(t)\ dt = \int_{t_0}^tv_0\ dt + \frac1m\int_{t_0}^t\int_{t_0}^t F(t)\ dt$$

$$x(t) = x_0 + v_0(t-t_0) + \frac1m\int_{t_0}^t\int_{t_0}^t F(t)\ dt$$

If $F$ is a constant, then we get back your original formula.