direct and inverse problems of mechanics

Motion of a material point is specified by the radius vector $\vec{r}(t)$; instantaneous velocity and acceleration are determined by derivatives: $\vec{v}(t)=\frac{d\vec{r}}{dt},\qquad \vec{a}(t)=\frac{d\vec{v}}{dt}=\frac{d^2\vec{r}}{dt^2}.$ In the direct problem of mechanics, the motion law $\vec{r}(t)$ is known, and one must find $\vec{v}(t)$ and $\vec{a}(t)$ by differentiation. In the inverse problem, acceleration $\vec{a}(t)$ (for example, from force) and initial conditions $\vec{r}(0)$ and $\vec{v}(0)$ are known, and one must reconstruct $\vec{v}(t)$ and $\vec{r}(t)$ by integration. Integration introduces constants that are fixed by the initial conditions: $\vec{v}(t)=\vec{v}(0)+\int_0^t \vec{a}(\tau)\,d\tau,\qquad \vec{r}(t)=\vec{r}(0)+\int_0^t \vec{v}(\tau)\,d\tau.$ Thus the direct and inverse problems are linked by differentiation and integration, and a solution is checked by the reverse operation.

A point moves according to $\vec{r}(t)=(2t^2)\,\hat{\imath}+(3t)\,\hat{\jmath}$ (in SI units). Find $\vec{v}(t)$ and $\vec{a}(t)$, then evaluate them at $t=2$ s.

A body is thrown vertically upward with initial speed $v_0=20$ m/s from height $y_0=0$. Its acceleration is $\vec{a}=-g\,\hat{\jmath}$, where $g=10$ m/s². Find $v(t)$ and $y(t)$, and the time to reach maximum height.

A particle has acceleration $\vec{a}(t)=(6t)\,\hat{\imath}-4\,\hat{\jmath}$ m/s². At $t=0$: $\vec{v}(0)=\hat{\imath}+2\,\hat{\jmath}$ m/s, $\vec{r}(0)=\vec{0}$. Find $\vec{v}(t)$ and $\vec{r}(t)$, and verify the result by differentiation.

Safe self-study plan: use an interactive simulation of projectile motion (for example, PhET) or a spreadsheet. Do not launch objects indoors. In the simulation, set initial $\vec{r}(0)$ and $\vec{v}(0)$, enable constant acceleration, and export a table of $x(t), y(t), v_x(t), v_y(t)$. Numerically differentiate the coordinates and compare with the given velocities; then numerically integrate the acceleration and compare with the given coordinates. Conclude which procedure solves the direct problem and which solves the inverse problem.

Show the relation between the direct and inverse problems: differentiation turns the motion law into velocity and acceleration, while integration with initial conditions returns velocity and position.

Flow diagram: three blocks $\vec{r}(t)\to\vec{v}(t)\to\vec{a}(t)$ are connected by arrows labeled “differentiate”; reverse arrows are labeled “integrate” and “+ initial conditions.” Beside it is a graph of $x(t), v_x(t), a_x(t)$ for the example $x(t)=2t^2$, showing that $v_x$ is the slope of $x(t)$ and $a_x$ is the slope of $v_x(t)$.

A car of mass $1200$ kg drives on dry asphalt at $20$ m/s. The brakes produce a constant force of $9000$ N directed against the motion. The force is known and the motion has to be found. The acceleration is $a=F/m=9000/1200=7{.}5$ m/s$^2$; the stopping time is $t=v_0/a=20/7{.}5\approx 2{.}7$ s; the braking distance is $s=v_0^2/(2a)=400/15\approx 27$ m.

This is the inverse problem of mechanics: Newton's second law $\vec a=\vec F/m$ gives the acceleration, and then, with $a$ constant, integrating it twice yields $v(t)=v_0-at$ and $x(t)=x_0+v_0t-\frac{a t^2}{2}$. These two formulas produce the numbers $t$ and $s$ above.

A smartphone strapped to a runner's arm logs the coordinate along the track every second: $0$, $2$, $6$, $12$ m. The average speeds over successive seconds are $2$, $4$ and $6$ m/s, so the speed grows by $2$ m/s each second. The runner's mass is $70$ kg, so the net force is $F=ma=70\cdot 2=140$ N.

This is the direct problem of mechanics: from the measured $x(t)$ one finds $v=\Delta x/\Delta t$, then $a=\Delta v/\Delta t$, and only then does Newton's second law turn the found acceleration into the force $F=ma$.

A ball of mass $0{.}43$ kg is kicked at $20$ m/s at $30^\circ$ to the ground; the contact lasts $0{.}02$ s. During the contact $a=20/0{.}02=1000$ m/s$^2$, so the force of the leg is $F=0{.}43\cdot 1000=430$ N. After that the ball flies under gravity alone: the flight time is $t=2v_0\sin\theta/g=2\cdot 20\cdot 0{.}5/9{.}8\approx 2{.}0$ s, the range is $R=v_0^2\sin 2\theta/g=400\cdot 0{.}866/9{.}8\approx 35$ m, and the highest point is $h=(v_0\sin\theta)^2/(2g)=100/19{.}6\approx 5{.}1$ m.

Both problems in one episode: the change of velocity during the kick is the direct problem, giving the force $F=m\Delta v/\Delta t$; the flight with constant acceleration $\vec a=-g\vec e_y$ is the inverse problem, where the components $v_x=v_0\cos\theta$ and $v_y=v_0\sin\theta-gt$ are integrated separately and give $t$, $R$ and $h$.