Kinematics with Vectors (DP IB Applications & Interpretation (AI): HL): Revision Note

Kinematics using vectors

What is kinematics?

• Kinematics is the use of mathematics to model motion in objects

• These are common terms used in kinematics:

  • Displacement describes the location of an object with respect to a fixed starting point

  • Velocity describes how the displacement changes over time

  • Acceleration describes how the velocity changes over time

• A vector for the velocity describes the direction an object is moving in

• The speed of an object is the magnitude of its velocity

How do I find whether two objects intersect?

• For example, consider two objects with position vectors:

  • ${\mathit{r}}_A=\left(\begin{array}{c}7\\ -1\\ -6\end{array}\right)+t\left(\begin{array}{c}-2\\ 1\\ 5\end{array}\right)$

  • ${\mathit{r}}_B=\left(\begin{array}{c}1\\ 5\\ 4\end{array}\right)+t\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$

• STEP 1
  Set the two equations equal to each other

  • $\left(\begin{array}{c}7\\ -1\\ -6\end{array}\right)+t\left(\begin{array}{c}-2\\ 1\\ 5\end{array}\right)=\left(\begin{array}{c}1\\ 5\\ 4\end{array}\right)+t\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$

• STEP2
  Form three equations using the components

  • $7-2t=1+t$

  • $-1+t=5-2t$

  • $-6+5t=4$

• STEP 3
  Solve any one of the equations

  • If the value is negative, then the objects do not intersect

    • $-6+5t=4\Rightarrow t=2$

• STEP 4
  Check whether that value of $t$ satisfies the other two equations

  • If it does, then the objects intersect

  • Otherwise, they do not

    • $7-2\left(2\right)=1+\left(2\right)=3$

    • $-1+\left(2\right)=5-2\left(2\right)=1$

    • Therefore, they intersect

• You can find the position vector of the point of intersection by substituting the value of $t$ back into one of the position vectors of the objects

How do I find the shortest distance between two moving objects?

• For example, consider two objects with position vectors:

  • ${\mathit{r}}_A=\left(\begin{array}{c}9\\ 0\\ -7\end{array}\right)+t\left(\begin{array}{c}-2\\ 1\\ 5\end{array}\right)$

  • ${\mathit{r}}_B=\left(\begin{array}{c}0\\ 6\\ -1\end{array}\right)+t\left(\begin{array}{c}1\\ -2\\ 0\end{array}\right)$

• STEP 1
  Find the displacement vector between the two position vectors

  • $\mathit{d}=\left(\begin{array}{c}9-2t\\ t\\ -7+5t\end{array}\right)-\left(\begin{array}{c}t\\ 6-2t\\ -1\end{array}\right)=\left(\begin{array}{c}9-3t\\ -6+3t\\ -6+5t\end{array}\right)$

• STEP 2
  Find the magnitude of this displacement vector

  • $\left|\mathit{d}\right|=\sqrt{{\left(9-3t\right)}^2+{\left(-6+3t\right)}^2+{\left(-6+5t\right)}^2}$

  • $\left|\mathit{d}\right|=\sqrt{43t^2-150t+153}$

• STEP 3
  Find the minimum value of the function under the square root

  • You can complete the square, use differentiation or use your GDC

    • Minimum of $43t^2-150t+153$ is $\frac{954}{43}$ and occurs when $t=\frac{75}{43}$

• STEP 4
  Take the positive square root of this value

  • $\sqrt{\frac{954}{43}}=4.710...$