Interpret successive rates
Position, velocity and acceleration
The top graph shows the position \(s(t) = t^3 - 9t^2 + 24t\), in meters, of a cart on a straight track after \(t\) seconds. It is green where the cart moves to the right and red where it moves to the left.
The middle graph shows the velocity in meters per second (m/s). The velocity is the slope of the position graph, so the orange bar is as tall as the slope of the orange tangent above it. The bottom graph shows the acceleration in meters per second per second (m/s²), which is the slope of the velocity graph. The middle graph is green where the velocity rises and red where it falls.
At \(t = 1\), the cart is at 16 m. Its velocity is 9 m/s, so it moves to the right. Its acceleration is \(-12\) m/s², so the velocity falls and the cart slows down.
Each rate is the slope of the one before
Each graph is the slope of the graph above it, so the velocity is \(v = s'\) and the acceleration is \(a = v' = s''\).
Each step divides by a time, so the unit has one more "per second".
The sign of the velocity gives the direction. The sign of the acceleration says whether the velocity rises or falls.
The speed is the size of the velocity. It increases when \(v\) and \(a\) have the same sign, and it decreases when they have opposite signs.
Notation
Velocity is the derivative of position with respect to time. Its unit is the unit of position divided by the unit of time, for example m/s.
\[\cy{v(t)} = s'(t) = \frac{ds}{dt}\]Acceleration is the derivative of velocity, so it is the second derivative of position. Its unit is m/s per second, written m/s².
\[\cy{a(t)} = v'(t) = s''(t) = \frac{d^2s}{dt^2}\]If \(v > 0\), the object moves in the positive direction, and the position graph is green. If \(v < 0\), it moves in the negative direction, and the graph is red.
If \(v\) and \(a\) have the same sign, the speed increases. If their signs differ, the speed decreases. The position function must be twice differentiable.
When it fails
At \(t = 3.5\) the acceleration is \(+3\) m/s², so it can seem that the cart speeds up. But the velocity is \(-2.25\) m/s, so the cart moves left.
The velocity rises from \(-3\) toward 0, so the cart moves left more and more slowly. The dashed curve is the speed, and it falls because \(v\) and \(a\) have opposite signs.
Example 1: a ball thrown upward
A ball leaves a roof at the height \(h(t) = -5t^2 + 20t + 1\) meters, where \(t\) is in seconds. Find its velocity and acceleration at \(t = 1\), with units.
Differentiate the height to find the velocity, in meters per second.
\[v(t) = h'(t) = -10t + 20\]Differentiate the velocity to find the acceleration, in meters per second per second.
\[a(t) = v'(t) = -10\]At \(t = 1\), the velocity is 10 m/s, so the ball moves up. The acceleration is \(-10\) m/s², and the signs differ, so the ball slows down.
\[v(1) = \cy{10}, \qquad a(1) = \cr{-10}\]At \(t = 2\), the velocity is 0 at the top of the path, where the height is 21 m.
\[v(2) = \cy{0}, \qquad h(2) = 21\]
Example 2: measured velocities
A cyclist starts at rest. Her velocity is measured every 5 seconds: 0, 6, 10 and 12 m/s at \(t = 0\), 5, 10 and 15. Find the average acceleration on each interval.
The graph joins the four measured points with straight segments. The average acceleration is the change in velocity divided by the change in time, so it is the slope of a segment, in m/s².
\[\begin{aligned} \frac{6 - 0}{5} &= 1.2 \\ \frac{10 - 6}{5} &= 0.8 \\ \frac{12 - 10}{5} &= 0.4 \end{aligned}\]A tempting wrong move: The accelerations fall from 1.2 to 0.8 to 0.4, so the cyclist slows down.
Every segment slopes up, so the velocity rises from 0 to 6 to 10 to 12, and the cyclist speeds up.
All three accelerations are positive, so the cyclist speeds up on each interval. The accelerations shrink, so each interval adds less speed than the one before.
Practice
A runner's position is \(s(t) = 2t^3 - 21t^2 + 60t\) feet, where \(t\) is in seconds. Find \(v(4)\) and \(a(4)\), with units. Does the runner speed up or slow down at \(t = 4\)?
Read the sign of the middle graph and the sign of the bottom graph at \(t = 4\).
Solution
Differentiate twice before you substitute.
\[\begin{aligned} v(t) &= 6t^2 - 42t + 60 \\ a(t) &= 12t - 42 \end{aligned}\]Substitute \(t = 4\). The velocity is \(-12\) ft/s and the acceleration is 6 ft/s².
\[\begin{aligned} v(4) &= 96 - 168 + 60 = \cr{-12} \\ a(4) &= 48 - 42 = \cg{6} \end{aligned}\]The velocity is negative, so the runner moves in the negative direction. The signs differ, so the runner slows down.
Questions
The figure shows the position, the velocity and the acceleration of a cart with \(s(t) = t^3 - 9t^2 + 24t\), where \(t\) is in seconds.
At \(t = 1\) the middle graph is above 0 and falling. The bottom graph is below 0. What is the cart doing?
- Moving right and slowing down
- Moving right and speeding up
- Moving left and slowing down
Answer
Moving right and slowing down
At \(t = 3.5\) the middle graph is below 0 and rising. What is the cart doing?
- Moving left and slowing down
- Moving left and speeding up
- Moving right and speeding up
Answer
Moving left and slowing down
A cart has the position \(s(t) = t^3 - 3t^2\) meters, where \(t\) is in seconds. What is its acceleration at \(t = 2\)?
- 6 m/s²
- 6 m/s
- 0 m/s²
- \(-4\) m
Answer
6 m/s²
Summary
You differentiate position to find velocity, and then you differentiate velocity to find acceleration, with the unit at each step.
This works because each quantity is the rate of change of the one before, so its unit has one more "per second".
You use the signs of velocity and acceleration to say which way an object moves and whether its speed rises or falls.
This is one of four skills in topic 3.6 of Which & Why: AP Calculus.