Find velocity and acceleration vectors
A thrown ball
A ball is thrown. After \(t\) seconds, its position in meters is \(\mathbf r(t) = \langle 20t,\ 30t - 5t^2\rangle\).
The orange arrow from the ball is the velocity \(\mathbf v(t) = \mathbf r'(t)\), and it points along the path. At \(t = 1\), it is \(\langle 20,\ 20\rangle\).
Drag the time. The horizontal part of the arrow stays 20. The vertical part is 20 at \(t = 1\), 10 at \(t = 2\), and 0 at \(t = 3\), the top of the path.
The vertical part falls by 10 in each second, so the velocity changes at the rate \(\langle 0,\ -10\rangle\). This rate of change of the velocity is the acceleration \(\mathbf a(t)\), the arrow with two heads.
The acceleration is the same at every time, but the velocity is not.
The derivative of the derivative
The velocity \(\mathbf v(t)\) is the derivative of the position vector, so its arrow shows how the position changes.
The acceleration \(\mathbf a(t)\) is the derivative of the velocity, so it shows how the velocity arrow changes. The faint arrows show the velocity one second apart, and each differs from the one before it by the arrow with two heads.
Drag the time. The velocity arrow changes, but in this motion the acceleration arrow stays the same.
The acceleration is the second derivative of the position, so you differentiate twice.
Notation
The position is a vector function.
\[\mathbf r(t) = \langle x(t),\ y(t)\rangle\]The velocity is the derivative of the position. You differentiate each component.
\[\mathbf v(t) = \mathbf r'(t) = \cy{\langle x'(t),\ y'(t)\rangle}\]The acceleration is the derivative of the velocity.
\[\begin{aligned} \mathbf a(t) &= \mathbf v'(t) = \mathbf r''(t) \\ &= \langle x''(t),\ y''(t)\rangle \end{aligned}\]The velocity is the first derivative of the position, and the acceleration is the second derivative.
This works only if each component has a second derivative at \(t\).
When it fails
Take \(\mathbf r(t) = \langle t,\ t|t|\rangle\). The velocity has the parts \(1\) and \(2|t|\), and it exists at every time, including \(t = 0\).
The acceleration is \(\langle 0,\ -2\rangle\) for \(t < 0\), pointing down, and \(\langle 0,\ 2\rangle\) for \(t > 0\), pointing up. The two sides differ at \(t = 0\), so \(\mathbf a(0)\) does not exist.
Example 1: Polynomial components
A point has position \(\mathbf r(t) = \langle t^3 - 3t,\ 2t^2\rangle\). Find its velocity and its acceleration.
The velocity is the derivative of the position. Differentiate each component.
\[\mathbf v(t) = \cy{\langle 3t^2 - 3,\ 4t\rangle}\]The acceleration is the derivative of the velocity. Differentiate the first component of \(\mathbf v\).
\[\dfrac{d}{dt}\left(3t^2 - 3\right) = 6t\]Differentiate the second component of \(\mathbf v\) and write the vector.
\[\mathbf a(t) = \langle 6t,\ 4\rangle\]At \(t = 1\), \(\mathbf v = \langle 0,\ 4\rangle\) points straight up and \(\mathbf a = \langle 6,\ 4\rangle\) points up and to the right. The figure draws both arrows at half their length.
Example 2: Motion on an ellipse
A point moves with \(\mathbf r(t) = \langle 5\cos t,\ 3\sin t\rangle\). Find \(\mathbf v(t)\) and \(\mathbf a(t)\).
Differentiate each component once.
\[\mathbf v(t) = \cy{\langle -5\sin t,\ 3\cos t\rangle}\]A tempting wrong move: The arrow along the curve is the acceleration.
\[\mathbf a(t) = \langle -5\sin t,\ 3\cos t\rangle\]At \(t = \frac{\pi}{2}\), this gives \(\langle -5,\ 0\rangle\), along the red line to the left. The faint arrow, the velocity a moment later, tilts down, so the velocity changes downward, not to the left. The arrow along the curve is the velocity, after only one derivative.
The acceleration is the derivative of the velocity. Differentiate each component again.
\[\mathbf a(t) = \langle -5\cos t,\ -3\sin t\rangle\]This is \(-\mathbf r(t)\), so the acceleration points from the point to the center of the ellipse. At \(t = \frac{\pi}{2}\), it is \(\langle 0,\ -3\rangle\), straight down.
Practice
A point has position \(\mathbf r(t) = \langle \tfrac32e^{-2t},\ 4t - \tfrac13t^3\rangle\). Find its velocity and its acceleration.
Drag the time. The point slows down and turns, so its velocity changes. Differentiate each component once for the velocity, and again for the acceleration.
Solution
Differentiate each component once. The first component needs the chain rule.
\[\mathbf v(t) = \cy{\left\langle -3e^{-2t},\ 4 - t^2\right\rangle}\]Differentiate each component of the velocity again.
\[\mathbf a(t) = \left\langle 6e^{-2t},\ -2t\right\rangle\]
Questions
Questions 1 and 2 use the curve \(\mathbf r(t) = \langle t^2,\ 6t - t^2\rangle\). The orange arrow is the velocity. The arrow with two heads is the acceleration.
At \(t = 2\), which vector is the acceleration?
- \(\langle 2,\ -2\rangle\)
- \(\langle 4,\ 2\rangle\)
- \(\langle 2,\ 2\rangle\)
Answer
\(\langle 2,\ -2\rangle\)
At \(t = 3\), the velocity arrow is horizontal. What is the acceleration?
- \(\langle 2,\ -2\rangle\)
- \(\langle 0,\ 0\rangle\)
- \(\langle 6,\ 0\rangle\)
Answer
\(\langle 2,\ -2\rangle\)
A point has position \(\mathbf r(t) = \langle \sin 3t,\ t^3\rangle\). What is its acceleration?
- \(\langle -9\sin 3t,\ 6t\rangle\)
- \(\langle 3\cos 3t,\ 3t^2\rangle\)
- \(\langle -3\sin 3t,\ 6t\rangle\)
- \(\langle -9\cos 3t,\ 6t\rangle\)
Answer
\(\langle -9\sin 3t,\ 6t\rangle\)
Summary
You find the velocity by differentiating each component of the position once, and the acceleration by differentiating each component again.
This works because the velocity is the rate of change of the position, and the acceleration is the rate of change of the velocity.
You use it when a position is given as a vector function of time.
This is one of four skills in topic 9.4 of Which & Why: AP Calculus.