The displacement vector [latex] \text{Δ}\overset{\to }{r} [/latex] gives the shortest distance between any two points on the trajectory of a particle in two or three dimensions. In fact, most of the time, instantaneous and average velocities are not the same. Here, we can notice that vx is treated as the adjacent side and vy as the opposite and v as the hypotenuse.
Calculate the average velocity in multiple dimensions. Calculate the velocity vector given the position vector as a function of time. The ball on the right has an initial horizontal velocity whereas the ball on the left has no horizontal velocity. Displacement and velocity in two or three dimensions are straightforward extensions of the one-dimensional definitions.
The typical problem will have some object, a boat or plane for example, which has a known velocity through some medium, air or water, which is itself in motion at a known speed. Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. The key to analyzing such motion, called projectile motion, is to resolve it into motions along perpendicular directions. Note that the satellite took a curved path along its circular orbit to get from its initial position to its final position in this example. ( The displacements in numerical order of a particle undergoing Brownian motion could look like the following, in micrometers (Figure \(\PageIndex{5}\)): \[\Delta \vec{r}_{1} = 2.0\; \hat{i} + \hat{j} + 3.0 \hat{k}\], \[\Delta \vec{r}_{2} = - \hat{i} + 3.0\; \hat{k}\], \[\Delta \vec{r}_{3} = 4.0\; \hat{i} -2.0\; \hat{j} + \hat{k}\], \[\Delta \vec{r}_{4} = -3.0\; \hat{i} + \hat{j} + 3.0\; \hat{k} \ldotp\]. The position function of a particle is \(\vec{r}\)(t) = 3.0t3 \(\hat{i}\) + 4.0 \(\hat{j}\). °
To find the vertical component of the velocity, we use the following relation. +
Now we can find \(\Delta \vec{r}\), the displacement of the satellite: \[\Delta \vec{r} = \vec{r} (t_{2}) - \vec{r} (t_{1}) = 4787\; \hat{i} - 11,557\; \hat{j} \ldotp \nonumber\], \[|\Delta \vec{r}| = \sqrt{(4787)^{2} + (-11,557)^{2}} = 12,509\; km. 10 [latex] \text{Δ}{\overset{\to }{r}}_{1}=20.00\,\text{m}\hat{j},\text{Δ}{\overset{\to }{r}}_{2}=(2.000\,×\,{10}^{4}\text{m})\,(\text{cos}30\text{°}\hat{i}+\text{sin}\,30\text{°}\hat{j}) [/latex], [latex] \text{Δ}\overset{\to }{r}=1.700\,×\,{10}^{4}\text{m}\hat{i}+1.002\,×\,{10}^{4}\text{m}\hat{j} [/latex], [latex] x=x(t)\quad y=y(t)\quad z=z(t).
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