Connect position, velocity, and acceleration
A cart on a track
A cart rolls along a straight track. The top graph shows its position \(s\), in meters from a mark on the track, after \(t\) seconds.
At \(t = 2\) the orange tangent has slope 8, so at that moment the cart moves at 8 meters per second. This rate is its velocity.
The middle graph plots this velocity \(v\). At \(t = 2\) its orange bar has height 8, which is the slope of the top graph.
The tangent on the middle graph has slope \(-2\), so the velocity falls by 2 meters per second each second. This is the acceleration \(a\), and the bottom graph plots it with a bar of height \(-2\).
At \(t = 7\) the middle bar is below the axis, so the cart moves back toward the mark.
Each graph is the slope of the one above
Without numbers, the graphs stand for any particle on a line with position \(s\).
Drag the point. At every time, the bar on each lower graph is as high as the slope of the graph above it.
Each step down is one derivative: velocity is the derivative of position, and acceleration is the derivative of velocity.
Notation
Velocity is the derivative of position, which is the rate at which position changes. The graph of \(s\) shows position against time, so it is not the path of the particle, which stays on the line.
\[\cy{v(t)} = s'(t) = \dfrac{ds}{dt}\]Acceleration is the derivative of velocity, which is the rate at which velocity changes. It is also the second derivative of position.
\[\cy{a(t)} = v'(t) = s''(t)\]A rate is a unit per unit of time. If s is in meters and t in seconds, v is in meters per second and a is in meters per second squared.
To find a value at a time, differentiate first and then substitute the time. If you substitute first, you have one number, and a number has no slope.
Example 1: A cyclist
A cyclist has position \(s(t) = 2t^2 + 3t\) meters after \(t\) seconds. Find the position, the velocity and the acceleration at \(t = 5\).
The position is the height of the point, so substitute 5 into s.
\[s(5) = 50 + 15 = \cy{65}\]The velocity is the derivative of s, which is the slope of the top graph.
\[v(t) = 4t + 3, \qquad v(5) = \cy{23}\]The acceleration is the derivative of v, the slope of the middle graph. This graph is a line with slope 4, so the acceleration is 4 at every time.
\[a(t) = 4, \qquad a(5) = \cy{4}\]At 5 seconds the cyclist is 65 meters from the mark and moves at 23 meters per second. The velocity grows by 4 meters per second each second.
Example 2: A cart with a cubic position
A cart has position \(s(t) = t^3 - 12t\) meters after \(t\) seconds. Find the velocity and the acceleration at \(t = 3\).
Substitute 3 into s to find where the cart is.
\[s(3) = 27 - 36 = \cy{-9}\]A tempting wrong move: The position at \(t = 3\) is the number \(-9\). The derivative of a number is 0, so the velocity is 0.
\[v(3) = \cr{0}\]The red line is flat, but the orange tangent at t = 3 rises steeply, so the slope is not 0. When you substitute first, s becomes one number, and a number has no slope.
Differentiate s first. Then substitute 3.
\[v(t) = 3t^2 - 12, \qquad v(3) = \cy{15}\]Differentiate v for the acceleration. Then substitute 3.
\[a(t) = 6t, \qquad a(3) = \cy{18}\]At 3 seconds the cart moves at 15 meters per second, and its velocity grows by 18 meters per second each second.
Practice
A bead on a wire has position \(s(t) = 4t^2 - t^3\) centimeters after \(t\) seconds. Find its velocity and its acceleration at \(t = 2\).
Differentiate \(s\) for the velocity, and differentiate the velocity for the acceleration. Then substitute 2.
Solution
Differentiate s. Then substitute 2.
\[v(t) = 8t - 3t^2, \qquad v(2) = \cy{4}\]Differentiate v. Then substitute 2.
\[a(t) = 8 - 6t, \qquad a(2) = \cy{-4}\]At 2 seconds the bead moves at 4 centimeters per second. Its velocity falls by 4 centimeters per second each second, because the slope of the velocity graph is negative.
Questions
Questions 1 and 2 use a particle with position \(s(t) = t^3 - 3t^2\) centimeters after \(t\) seconds. The three graphs show \(s\), \(v\) and \(a\).
At \(t = 1\) the acceleration is 0 and the velocity is \(-3\). Which statement is correct?
- The particle moves in the negative direction at 3 centimeters per second, and its velocity is not changing at that moment.
- The particle is at rest, because the acceleration is 0.
- The particle moves in the positive direction at 3 centimeters per second.
- The particle is 3 centimeters from the mark, on the negative side.
Answer
The particle moves in the negative direction at 3 centimeters per second, and its velocity is not changing at that moment.
At \(t = 3\) the tangent on the middle graph has slope 12. Which quantity equals 12?
- The acceleration, in centimeters per second squared
- The velocity, in centimeters per second
- The position, in centimeters
Answer
The acceleration, in centimeters per second squared
A particle has position \(s(t) = t^4 - 8t\) centimeters. What is its acceleration at \(t = 3\)?
- 108 centimeters per second squared
- 100 centimeters per second squared
- 57 centimeters per second squared
- 36 centimeters per second squared
Answer
108 centimeters per second squared
Summary
The velocity is the derivative of the position, and the acceleration is the derivative of the velocity.
Each graph is the slope of the one above it, so each step down is one derivative.
When a problem gives a position and asks for a velocity or an acceleration, you differentiate first and then substitute the time.
This is one of five skills in topic 4.2 of Which & Why: AP Calculus.