Menu
Sign up Sign in

AI HL · AHL 4

Vectors

0 of 160 skills done

0
AHL 4.27

An object that starts later

AHL 4.27

An object that starts later

Target: I can write the displacement, velocity or position of an object that starts later, by replacing \(t\) with \((t - a)\). (+2 more)

By the end of this section

Key points

An object that starts later

  • If an object starts at \(t = a\) instead of \(t = 0\), replace \(t\) by \((t - a)\) in its displacement, velocity and acceleration. This is true for \(t \ge a\).
  • Replace \(t\) everywhere it appears, including inside \(t^2\). For constant velocity, \(\mathbf{r} = \mathbf{r}_0 + (t - a)\mathbf{v}\).
  • The graph is translated \(a\) units to the right. This is the same idea as replacing \(x\) by \((x - a)\) in a function.

Comparing two objects

  • Keep one clock. Time \(t\) is measured from the start of the first object. The time since the second object started is \(t - a\).
  • \(1\) second after the second object starts is \(t = a + 1\). Put this \(t\) into both displacements.
  • The distance between them is \(|\mathbf{r}_2 - \mathbf{r}_1|\), the magnitude of the difference.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
The distance between two stones

A stone is thrown from the origin and its displacement at time \(t\) seconds is \(\mathbf{r}_1 = \begin{pmatrix} 4t \\ 10t - 4.9t^2 \end{pmatrix}\) m, where the first component is horizontal and the second is upward. A second stone is thrown along the same path 0.4 seconds later. Find the distance between the two stones 1.2 seconds after the second stone is thrown.

Working
  1. Second stone

    \(\mathbf{r}_2 = \begin{pmatrix} 4(t - 0.4) \\ 10(t - 0.4) - 4.9(t - 0.4)^2 \end{pmatrix}\)

    It starts when \(t = 0.4\). Replace \(t\) by \((t - 0.4)\) in every place.

  2. The clock time

    1.2 s after the second throw: \(t = 0.4 + 1.2 = 1.6\)

    The first stone has been moving for 1.6 s.

  3. Both positions

    \(\mathbf{r}_1 = \begin{pmatrix} 6.4 \\ 3.456 \end{pmatrix},\ \mathbf{r}_2 = \begin{pmatrix} 4.8 \\ 4.944 \end{pmatrix}\)

    Put \(t = 1.6\) into \(\mathbf{r}_1\), and \(t = 1.6\) into \(\mathbf{r}_2\). That is \((1.6 - 0.4) = 1.2\).

  4. Difference

    \(\mathbf{r}_2 - \mathbf{r}_1 = \begin{pmatrix} -1.6 \\ 1.488 \end{pmatrix}\)

    Subtract the first position from the second.

  5. Distance

    \(\sqrt{(-1.6)^2 + (1.488)^2} = 2.18\) m

    The distance is the magnitude of the difference.

Video walkthrough coming
Writing the displacement of a later object

A ball is thrown from the origin and its displacement at time \(t\) seconds is \(\mathbf{r}_1 = \begin{pmatrix} 3t \\ 8t - 4.9t^2 \end{pmatrix}\) m. A second ball is thrown from the origin along the same path 2 seconds later. (a) Write down the displacement of the second ball at time \(t\). (b) Find the height of the second ball when \(t = 3\).

Working
  1. Replace t

    \(\mathbf{r}_2 = \begin{pmatrix} 3(t - 2) \\ 8(t - 2) - 4.9(t - 2)^2 \end{pmatrix}\)

    The second ball starts at \(t = 2\), so replace every \(t\) by \((t - 2)\), including the one in \(t^2\). This is for \(t \ge 2\).

  2. Put in t = 3

    \(t - 2 = 1\), \(\mathbf{r}_2 = \begin{pmatrix} 3(1) \\ 8(1) - 4.9(1)^2 \end{pmatrix} = \begin{pmatrix} 3 \\ 3.1 \end{pmatrix}\)

    At clock time 3 the second ball has been in the air for 1 second.

  3. Height

    3.1 m

    The height is the second component.

Video walkthrough coming
Two drones, one starting later

Drone A starts at time \(t = 0\) seconds. Its position vector in metres is \(\mathbf{r}_A = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} + t\begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix}\). Drone B starts 2 seconds later from the point \((10, 5, 3)\) with velocity \(\begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}\) m s⁻¹. Find the distance between the drones when \(t = 5\).

Working
  1. Drone B

    \(\mathbf{r}_B = \begin{pmatrix} 10 \\ 5 \\ 3 \end{pmatrix} + (t - 2)\begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}\)

    It starts at the point when \(t = 2\): replace \(t\) by \((t - 2)\).

  2. Drone A at t = 5

    \(\mathbf{r}_A = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} + 5\begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 21 \\ 13 \\ 5 \end{pmatrix}\)

    Substitute \(t = 5\).

  3. Drone B at t = 5

    \(\mathbf{r}_B = \begin{pmatrix} 10 \\ 5 \\ 3 \end{pmatrix} + 3\begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 8 \\ 9 \end{pmatrix}\)

    Here \(t - 2 = 3\).

  4. Difference

    \(\mathbf{r}_B - \mathbf{r}_A = \begin{pmatrix} -17 \\ -5 \\ 4 \end{pmatrix}\)

    Subtract.

  5. Distance

    \(\sqrt{(-17)^2 + (-5)^2 + (4)^2} = 18.2\) m

    The distance is the magnitude of the difference.

Video walkthrough coming