​​Vectors in kinematics

​​In a nutshell

Position vectors are quantities that determine the position of one point in space, always relative to another point. Vectors are useful tools used to describe motion in a plane, as they have a direction and a magnitude.

Equations

Variable definitions

Vectors in kinematics

PROCEDURE

Example 1

A particle starts from the point with position vector $$(5\bold{i} + 10 \bold{j}) \ m$$ and moves with constant velocity $$(2\bold{i} - 2 \bold{j})\ ms^{-1}$$.

i) Find the position vector of this particle after $$10$$ seconds.

ii) Show that the particle is never at the point $$(0,0).$$

i) Find the position vector of this particle after $$10$$ seconds. 

Using the equation $$\bold r = \bold r_0 + \bold vt$$, substitute $$\bold r_0 =(5\bold{i} + 10 \bold{j})\, m$$ and $$\bold v=(2\bold{i} - 2 \bold{j})\ ms^{-1}$$:

$$\bold r = \bold r_0 + \bold v t = \left[(5\bold i + 10 \bold{j}) +(2\bold{i} - 2 \bold{j})t\right] \ m$$​​

Group the terms according to the unit vectors: 

​$$\bold r = \left[(5+2t)\bold{i} +(10-2t)\bold j\right]\ m$$​

Now substitute the value of the time $$t=10 \ s$$:

​$$\begin{aligned}\bold r &= (5+2\times 10)\bold{i} +(10-2\times 10)\bold j \\&=(5+20)\bold{i} +(10-20)\bold j \\&= (25\bold{i} -10\bold j) \ m\end{aligned}$$​​

In conclusion, $$\underline{\bold r = (25 \bold i - 10 \bold j) \ m}$$.

ii) Show that the particle is never at the point $$(0,0).$$

Using $$\bold r = \left[(5+2t)\bold{i} +(10-2t)\bold j\right]m$$ (from the first part), you only have to substitute $$\bold r = (0 \bold i + 0 \bold j) \ m$$. Therefore:

$$\bold r = 0\bold i + 0 \bold j = (5+2t)\bold{i} +(10-2t)\bold j\implies\begin{cases}0 \bold i = (5 +2t)\bold i \\0 \bold j = (10 -2 t )\bold j\end{cases} \implies \begin{cases}t=-0.4 \ s\\t = 5 \ s\end{cases}$$​​

The particle is never at the origin because the two times are not consistent.

​

Example 2

A particle $$P$$ initially moves with velocity $$(3\bold i - 8\bold j)\ ms^{-1}$$. It has an acceleration of $$(-2 \bold i + 3 \bold j) \ ms^{-2}$$.

i) Find the velocity of $$P$$ after $$t=3 \ s.$$​

ii) Find the speed of $$P$$ at $$t = 3 \ s$$ to $$2\space d.p.$$​

i) Find the velocity of $$P$$ after $$t=3 \ s.$$​

Using the equation $$\bold v=\bold u +\bold a t$$ you have to substitute $$\bold u =(3\bold i - 8\bold j)\ ms^{-1}$$ and $$\bold a=(-2 \bold i + 3 \bold j) \ ms^{-2}$$. Therefore:

$$\bold v = \left[(3\bold i - 8 \bold j) + (-2 \bold i + 3 \bold j)t\right]\ ms^{-1}$$​​

Group the terms according to the unit vectors.

$$\bold v = \left[(3 - 2t)\bold i + (-8 + 3t)\bold j\right] \ ms^{-1}$$​​

Now substitute the value of the time $$t = 3 \ s:$$​

$$\begin{aligned}\bold v &= (3 - 2\times 3)\bold i + (-8 + 3\times 3)\bold j \\&=(3 -6)\bold i + (-8 + 9)\bold j \\&= (-3\bold i + \bold j) \ ms^{-1}\end{aligned}$$​​

In conclusion $$\underline{\bold v = (-3 \bold i +\bold j) \ ms^{-1}}$$​

ii) Find the speed of $$P$$​ at $$t = 3 \ s.$$

The formula for speed is $$\sqrt{v_x^2 + v_y^2}$$, therefore the speed of $$P$$ is:

$$Speed \ of \ P = \sqrt{(-3)^2 +1^2 }=\sqrt{9+ 1}=\sqrt{10} = \underline{3.16 \ ms^{-1} \ (2 \ d.p.)}$$​​