Wemazi

Topic 9.6 · BC

Connect position, velocity, and acceleration

Builds on distinguishing displacement from distance. 7 minutes.

A thrown ball

A ball is thrown. After \(t\) seconds, its position in meters is \(\langle 3t,\ 8t - 2t^2\rangle\).

The orange arrow from the ball is the velocity, which points along the path and shows how fast the position changes.

At \(t = 1\), the velocity is \(\langle 3,\ 4\rangle\). At \(t = 2\), it is \(\langle 3,\ 0\rangle\), so the ball moves horizontally for an instant. At \(t = 3\), it is \(\langle 3,\ -4\rangle\).

Press the last button to see the three arrows together. In each second, the velocity changes by \(\langle 0,\ -4\rangle\). The arrow with two heads shows this rate of change, which is the acceleration.

The figure draws both arrows at half their length, so that they fit.

One derivative at a time

Drag the point. The orange arrow is the velocity, and it points along the path. Press Show a to add the acceleration, the arrow with two heads, which shows how the orange arrow changes.

The three boxes under the plane are the position r, the velocity v and the acceleration a. The lit box names the vector that the plane shows.

Each quantity is the rate of change of the one before it, so a step to the right is one derivative of each component. A step to the left is one integration, which adds a constant vector because the derivative of a constant is 0.

Notation

  1. The position is a vector function of time.

    \[\mathbf r(t) = \langle x(t),\ y(t)\rangle\]

  2. The velocity is the derivative of the position. You differentiate each component.

    \[\cy{\mathbf v}(t) = \mathbf r'(t) = \langle x'(t),\ y'(t)\rangle\]

  3. The acceleration is the derivative of the velocity, so it is the second derivative of the position.

    \[\begin{aligned}\mathbf a(t) &= \mathbf v'(t) = \mathbf r''(t) \\ &= \langle x''(t),\ y''(t)\rangle\end{aligned}\]

  4. To go back, you integrate and add the value at a starting time \(t_0\). Each integration needs one starting value.

    \[\begin{aligned}\mathbf v(t) &= \mathbf v(t_0) + \int_{t_0}^{t} \mathbf a(u)\,du, \\ \mathbf r(t) &= \mathbf r(t_0) + \int_{t_0}^{t} \mathbf v(u)\,du\end{aligned}\]

When it fails

The second ball follows the first path shifted by \(\langle 2,\ 1\rangle\). At every time, its orange arrow is the same as the arrow of the first ball.

Because the two balls have the same velocity, the velocity alone cannot give the position. You need the position at one time to choose the path.

Example 1: Differentiate twice

A particle has position \(\mathbf r(t) = \langle t^3 - 3t,\ 2t^2 - t\rangle\). Find its velocity and acceleration at \(t = 2\).

  1. Differentiate each component of the position to find the velocity.

    \[\mathbf v(t) = \cy{\langle 3t^2 - 3,\ 4t - 1\rangle}\]

  2. Differentiate each component of the velocity to find the acceleration.

    \[\mathbf a(t) = \langle 6t,\ 4\rangle\]

  3. Substitute \(t = 2\) into both.

    \[\mathbf v(2) = \cy{\langle 9,\ 7\rangle},\qquad \mathbf a(2) = \langle 12,\ 4\rangle\]

    The figure draws both arrows shorter than they are, so that they fit.

Example 2: Integrate twice

A particle has acceleration \(\mathbf a(t) = \langle 2,\ 12t\rangle\). At \(t = 0\), its velocity is \(\langle 4,\ 0\rangle\) and its position is \((0,\ 3)\). Find its position at \(t = 1\).

  1. Integrate each component of the acceleration. At \(t = 0\), the terms with \(t\) are 0, so the constants are the components of the starting velocity, 4 and 0.

    \[\mathbf v(t) = \cy{\langle 2t + 4,\ 6t^2\rangle}\]

  2. Integrate each component of the velocity. This integration adds two new constants.

    \[\mathbf r(t) = \langle t^2 + 4t + C_1,\ 2t^3 + C_2\rangle\]

  3. A tempting wrong move: Use the starting velocity \(\langle 4,\ 0\rangle\) again for the constants.

    \[\mathbf r(t) = \langle t^2 + 4t + 4,\ 2t^3\rangle\]

    The red path shows this position, and it starts at \((4,\ 0)\) instead of at \((0,\ 3)\). A starting velocity is not a starting position.

  4. Use the starting position \((0,\ 3)\) instead. At \(t = 0\), the terms with \(t\) are 0, so \(C_1 = 0\) and \(C_2 = 3\). Then substitute \(t = 1\).

    \[\begin{aligned}\mathbf r(t) &= \langle t^2 + 4t,\ 2t^3 + 3\rangle, \\ \mathbf r(1) &= \cy{\langle 5,\ 5\rangle}\end{aligned}\]

Practice

A particle has position \(\mathbf r(t) = \langle 2\sin t,\ t^2 - 2t\rangle\). Find its velocity and acceleration at \(t = \pi/2\). Then say which way the particle moves at that time.

Drag the slider to \(t = 1.57\), which is close to \(\pi/2\). The direction of the path there is the direction of the velocity. Differentiate each component once for the velocity and twice for the acceleration.

Solution

  1. Differentiate each component twice.

    \[\begin{aligned}\mathbf v(t) &= \langle 2\cos t,\ 2t - 2\rangle, \\ \mathbf a(t) &= \langle -2\sin t,\ 2\rangle\end{aligned}\]

  2. Substitute \(t = \pi/2\).

    \[\begin{aligned}\mathbf v\!\left(\tfrac{\pi}{2}\right) &= \cy{\langle 0,\ \pi - 2\rangle}, \\ \mathbf a\!\left(\tfrac{\pi}{2}\right) &= \langle -2,\ 2\rangle\end{aligned}\]

  3. The x-component of the velocity is 0, and the y-component is positive because \(\pi - 2 \approx 1.14\), so the particle moves straight up.

Questions

A particle has position \(\langle 4t - t^2,\ 2t\rangle\) for \(0 \le t \le 3\). The orange arrow shows the velocity at \(t = 1\). The faint arrow shows the velocity at \(t = 2\).

  1. The velocity changes from \(\langle 2,\ 2\rangle\) at \(t = 1\) to \(\langle 0,\ 2\rangle\) at \(t = 2\). What is the acceleration?

    • \(\langle -2,\ 0\rangle\)
    • \(\langle 0,\ 2\rangle\)
    • \(\langle 2,\ 0\rangle\)
    Answer

    \(\langle -2,\ 0\rangle\)

  2. At \(t = 2\), the particle is at the rightmost point of its path. What is its velocity?

    • \(\langle 0,\ 2\rangle\)
    • \(\langle 0,\ 0\rangle\)
    • \(\langle 4,\ 4\rangle\)
    Answer

    \(\langle 0,\ 2\rangle\)

  3. A particle has acceleration \(\langle 6t,\ -2\rangle\), and its velocity at \(t = 0\) is \(\langle 1,\ 0\rangle\). What is its velocity at \(t = 1\)?

    • \(\langle 4,\ -2\rangle\)
    • \(\langle 3,\ -2\rangle\)
    • \(\langle 6,\ -2\rangle\)
    Answer

    \(\langle 4,\ -2\rangle\)

Summary

You differentiate the position once to find the velocity and twice to find the acceleration, one component at a time.

To go back, you integrate and use one starting value for each integration, because a derivative does not show the constant that each integration adds.

You use it for any motion in the plane when you know one of the three vectors.

This is one of five skills in topic 9.6 of Which & Why: AP Calculus.