Vectors projections and displacement

Displacement is the vector connecting the initial and final positions of a body: $\vec{s} = \vec{r}_2 - \vec{r}_1$. The projection of a vector onto an axis is a signed number equal to the coordinate shift along that axis: $s_x = x_2 - x_1$ and $s_y = y_2 - y_1$. The sign of a projection gives the direction along the axis: plus means along the axis, minus means opposite to it, and the magnitude of the vector follows from the Pythagorean theorem, $s = \sqrt{s_x^2 + s_y^2}$. A motion graph links the projections to time: on an $x(t)$ graph the displacement projection is the difference of ordinates, while on a $v_x(t)$ graph it is the signed area under the curve. Therefore distance travelled and the magnitude of displacement coincide only when the body does not reverse its direction of motion. This is how a graph becomes the source of input data for a vector calculation.

On an $x(t)$ graph a body has coordinate $x_1 = 2$ m at $t_1 = 0$ and coordinate $x_2 = 8$ m at $t_2 = 4$ s. Find the projection of the displacement on the $Ox$ axis over this time interval.

A point moves from position $A(1;\,2)$ m to position $B(4;\,6)$ m. Find the projections of the displacement on the $Ox$ and $Oy$ axes, the magnitude of the displacement, and the angle between the displacement vector and the $Ox$ axis.

On a $v_x(t)$ graph the velocity projection is $v_x = +4$ m/s from $t = 0$ to $t = 3$ s, and then $v_x = -2$ m/s from $t = 3$ s to $t = 5$ s. Find the displacement projection on the $Ox$ axis over the whole 5 s, the magnitude of displacement, and the distance travelled.

Work in a simulation (for example PhET "Vector Addition" or "Moving Man") so that you do not measure displacements on roads or in hazardous places.

1. Set the initial point $A(1;\,1)$ m and the final point $B(5;\,4)$ m; switch on the display of vector components.
2. Compare the components shown by the simulation with your calculation $s_x = x_2 - x_1$, $s_y = y_2 - y_1$ and the magnitude $s = \sqrt{s_x^2 + s_y^2}$.
3. Move the points so that one projection becomes negative and explain how this changed the sign and the direction of the vector.
4. Build a $v_x(t)$ graph from two horizontal segments of opposite sign and check that the signed area gives the same displacement projection as the difference of coordinates.
5. Record the results in a table: point $A$, point $B$, $s_x$, $s_y$, $s$, angle, distance.

Show how one and the same displacement vector splits into projections onto the axes, and how the sign of a projection is read off a motion graph.

Sketch: axes $Ox$ and $Oy$, points $A$ and $B$, the vector $\vec{s} = \overrightarrow{AB}$, dashed perpendiculars to the axes labelled $s_x = x_2 - x_1$ and $s_y = y_2 - y_1$, and a right triangle with hypotenuse $s$ and angle $\theta$ at the $Ox$ axis. Beside it, a $v_x(t)$ graph made of two horizontal segments with the shaded areas $+12$ m and $-4$ m.

A footballer takes the ball at the corner flag and runs to meet a pass: first $30$ m along the touchline (east), then $40$ m up the pitch (north). The coach asks how far from the corner flag and in which direction the player ends up.

Displacement is the vector sum of the two legs: projections onto the east–north axes add component-wise ($s_x=30$ m, $s_y=40$ m), the magnitude follows from Pythagoras $|\vec s|=\sqrt{s_x^2+s_y^2}=50$ m, and the direction is set by $\tan\theta=s_y/s_x$, i.e. $\theta\approx 53^\circ$ from east.

An aircraft flies at $200$ km/h due east while a crosswind pushes it north at $50$ km/h. The controller needs the aircraft's speed relative to the ground and the drift angle.

Ground speed is the vector sum of the aircraft's own velocity and the wind velocity: the projections onto the axes are $v_x=200$ km/h and $v_y=50$ km/h, the magnitude is $|\vec v|=\sqrt{200^2+50^2}\approx 206$ km/h, and the drift angle $\arctan(50/200)\approx 14^\circ$ follows from the ratio of the projections.

A cable of a suspension bridge is tensioned with a force of $5000$ N at $30^\circ$ to the horizontal and holds part of the span. The engineer needs to know which part of the tension pulls the span horizontally and which part vertically.

Force is a vector, and its projections onto the axes come from the cosine and sine of the angle: $F_x=F\cos 30^\circ\approx 4330$ N and $F_y=F\sin 30^\circ=2500$ N. These projections are independent, and they show which shares of one and the same vector act along each direction.