Constant Force = Parabolic Motion

When the total force acting on a particle is constant, $\vec{F}=F_x\,\hat{x}+F_y\,\hat{y}+F_z\,\hat{z}$, the acceleration is also constant,

$$\vec{a}=\frac{F_x}{m}\,\hat{x}+\frac{F_y}{m}\,\hat{y}+\frac{F_z}{m}\,\hat{z}.$$

The velocity vector can be determined by integrating each component of the acceleration vector. For the $x$-component, $v_x=\int a_xdt=\frac{F_xt}{m}+C$. The integration constant $C$ can be determined by considering time $t=0$. At $t=0$ the term $F_xt/m=0$ and the integration constant is the $x$-component of the velocity at $t=0$, $C=v_{x0}$. Integrating the $y$- and $z$-components similarly yields the velocity vector,

$$\vec{v}=\left(v_{x0}+\frac{F_x}{m}t\right)\,\hat{x}+\left(v_{y0}+\frac{F_y}{m}t\right)\,\hat{y}+\left(v_{z0}+\frac{F_z}{m}t\right)\,\hat{z}.$$

The position vector can be determined by integrating each component of the velocity vector. For the $x$-component, $x=\int v_xdt=v_{x0}t+\frac{F_xt^2}{2m}+C$. The integration constant $C$ can be determined by considering time $t=0$. At $t=0$ the terms $v_{x0}t+\frac{F_xt^2}{2m}=0$ and the integration constant is the $x$-component of the position vector at $t=0$, $C=x_{0}$. Integrating the $y$- and $z$-components similarly yields the position vector,

$$\vec{r}=\left(x_0 +v_{x0}t+\frac{F_x}{2m}t^2\right)\,\hat{x}+\left(y_0 +v_{y0}t+\frac{F_y}{2m}t^2\right)\,\hat{y}+\left(z_0 +v_{z0}t+\frac{F_z}{2m}t^2\right)\,\hat{z}$$

When a particle experiences a constant force, each component of the position vector is a parabolic function of time.

This problem can be solved numerically using the <a href="http://lampz.tugraz.at/~hadley/physikm/apps/numerical_integration/ode2ball.en.php">Numerical differential equation solver</a>.






	Physik M 513.805 (511.015)