This note shows how calculus describes motion in a straight line. You will link displacement, velocity and acceleration by differentiating and integrating, and separate displacement from distance travelled. These ideas turn up whenever a question gives you a formula for the position or speed of a moving object.
Key ideas
Take a particle moving along a straight line. Its position at time \(t\) is the displacement \(s\) from a fixed origin. Displacement can be positive or negative, depending on which side of the origin the particle is on.
- Velocity is the rate of change of displacement: \(v=\dfrac{ds}{dt}\).
- Acceleration is the rate of change of velocity: \(a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}\).
- Going the other way, integrate: \(v=\int a\,dt\) and \(s=\int v\,dt\). Each indefinite integral brings a constant, which you find from given information such as \(s=2\) when \(t=0\).
Velocity has a sign. A positive velocity means moving in the positive direction, a negative velocity means moving in the negative direction. Speed is the size of the velocity, \(|v|\), so it is never negative.
Key moments. The particle is momentarily at rest when \(v=0\). It changes direction when \(v\) changes sign, which is usually at a point where \(v=0\). The particle is at the origin when \(s=0\). Maximum or minimum displacement happens where \(v=0\) and the sign of \(v\) changes. Maximum or minimum velocity happens where \(a=0\) and the sign of \(a\) changes.
Speeding up or slowing down. Compare the signs of \(v\) and \(a\). If they are the same, the speed is increasing. If they are different, the speed is decreasing. This is about speed, so a negative acceleration does not always mean slowing down.
Displacement and distance. Over the time interval \(t_1\le t\le t_2\):
- Change in displacement \(=\displaystyle\int_{t_1}^{t_2} v\,dt\). Areas below the \(t\)-axis count as negative.
- Total distance travelled \(=\displaystyle\int_{t_1}^{t_2} |v|\,dt\). Every area counts as positive.
To find the distance by hand, find where \(v=0\) inside the interval, split the interval there, integrate each part and add the sizes. On your GDC, you can enter \(|v|\) directly and use the definite integral feature, or sketch the graph of \(v\) to see where it crosses the axis.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(v=\dfrac{ds}{dt}\) | Velocity from displacement. Differentiate. |
| \(a=\dfrac{dv}{dt}\) | Acceleration from velocity. Differentiate. |
| \(s=\int v\,dt\) | Displacement from velocity. Use a given condition to find the constant. |
| \(v=\int a\,dt\) | Velocity from acceleration. Use a given condition to find the constant. |
| Speed \(=|v|\) | Size of the velocity. Used when asked how fast, not which way. |
| \(\displaystyle\int_{t_1}^{t_2} v\,dt\) | Change in displacement (signed). |
| \(\displaystyle\int_{t_1}^{t_2} |v|\,dt\) | Total distance travelled. |
| \(v=0\) | Particle at rest. Often a turning point. |
Worked examples
Example 1 A robot arm moves along a rail. Its displacement from the start of the rail after \(t\) seconds is \(s=t^3-6t^2+9t\) metres, for \(t\ge 0\). (a) Find the times when the arm is at rest. (b) Find the acceleration at the second of those times. (c) Find the total distance travelled in the first 4 seconds.
(a) Differentiate: \(v=\dfrac{ds}{dt}=3t^2-12t+9=3(t-1)(t-3)\).
At rest means \(v=0\), so \(t=1\) or \(t=3\).
(b) Differentiate again: \(a=\dfrac{dv}{dt}=6t-12\). At \(t=3\), \(a=6(3)-12=6\).
(c) The arm changes direction at \(t=1\) and \(t=3\), so split the 4 seconds into three parts. Find the displacement at each key time.
\(s(0)=0\). \(s(1)=1-6+9=4\). \(s(3)=27-54+27=0\). \(s(4)=64-96+36=4\).
From 0 to 1 the arm moves 4 m forward. From 1 to 3 it moves 4 m back. From 3 to 4 it moves 4 m forward.
Total distance \(=4+4+4=12\) m.
Answer: (a) \(t=1\) s and \(t=3\) s; (b) \(6\ \text{m s}^{-2}\); (c) 12 m (the final displacement is only 4 m).
Example 2 A sled on a straight track has velocity \(v=6t-3t^2\) m s\(^{-1}\) and is at \(s=2\) m when \(t=0\). (a) Find an expression for \(s\). (b) Find the greatest value of \(s\) for \(0\le t\le 3\). (c) Find the distance travelled between \(t=0\) and \(t=3\).
(a) Integrate: \(s=\int(6t-3t^2)\,dt=3t^2-t^3+C\).
When \(t=0\), \(s=2\), so \(C=2\). Therefore \(s=3t^2-t^3+2\).
(b) Write \(v=3t(2-t)\). Then \(v=0\) at \(t=0\) and \(t=2\). For \(0\lt t\lt 2\), \(v\gt 0\), and for \(t\gt 2\), \(v\lt 0\). So the sled moves forward until \(t=2\) and then comes back.
\(s(2)=12-8+2=6\). The greatest displacement in this interval is 6 m (the value at \(t=2\); at \(t=0\) and \(t=3\) it is 2 m).
(c) \(s(3)=27-27+2=2\). From \(t=0\) to \(t=2\) the sled moves \(6-2=4\) m forward. From \(t=2\) to \(t=3\) it moves \(6-2=4\) m back.
Distance \(=4+4=8\) m.
Answer: (a) \(s=3t^2-t^3+2\); (b) 6 m; (c) 8 m.
Example 3 A lift cage has velocity \(v=t^2-4t\) m s\(^{-1}\) (upwards positive). Decide whether its speed is increasing or decreasing at (a) \(t=1\) and (b) \(t=3\).
The acceleration is \(a=\dfrac{dv}{dt}=2t-4\).
(a) At \(t=1\): \(v=1-4=-3\) and \(a=2-4=-2\). Both are negative, so they have the same sign. The speed is 3 m s\(^{-1}\) and increasing.
(b) At \(t=3\): \(v=9-12=-3\) and \(a=6-4=2\). The signs differ, so the speed is decreasing.
Answer: speeding up at \(t=1\); slowing down at \(t=3\).
Common errors
- Common error: giving the change in displacement when the question asks for distance. If the particle turns round, \(\int v\,dt\) is smaller than the distance, because areas below the axis cancel.
- Common error: saying a particle is slowing down because \(a\) is negative. Compare the signs of \(v\) and \(a\) instead.
- Common error: forgetting the constant of integration, then not using the starting condition.
- Common error: treating \(v=0\) as the answer to "when is the particle at the origin". That is \(s=0\).
- Common error: dropping units, or writing a negative speed. Speed is \(|v|\); the sign belongs to velocity.
Exam tip: For a distance question, sketch a quick graph of \(v\) against \(t\) or make a small table of \(s\) at the turning times. It shows you at once where to split the interval.
Check yourself
- A cyclist's displacement is \(s=2t^3-9t^2+12t\) metres. Find the times when the cyclist is at rest.
Answer
\(v=6t^2-18t+12=6(t-1)(t-2)\). So \(v=0\) when \(t=1\) s and \(t=2\) s. - A drone has acceleration \(a=6t-4\) m s\(^{-2}\), with \(v=1\) m s\(^{-1}\) and \(s=0\) when \(t=0\). Find \(v\) and \(s\) when \(t=2\).
Answer
\(v=3t^2-4t+C\), and \(v(0)=1\) gives \(C=1\), so \(v=3t^2-4t+1\). Then \(s=t^3-2t^2+t+D\) with \(D=0\). At \(t=2\): \(v=12-8+1=5\) m s\(^{-1}\) and \(s=8-8+2=2\) m. - A particle has \(v=t^2-6t+8\) m s\(^{-1}\). Find the displacement and the distance travelled from \(t=0\) to \(t=4\).
Answer
\(v=(t-2)(t-4)\), positive for \(t\lt 2\) and negative for \(2\lt t\lt 4\). With \(F(t)=\frac{t^3}{3}-3t^2+8t\): \(F(2)=\frac{20}{3}\), \(F(4)=\frac{16}{3}\). Displacement \(=\frac{16}{3}\approx 5.33\) m. Distance \(=\frac{20}{3}+\left(\frac{20}{3}-\frac{16}{3}\right)=8\) m. - A boat has velocity \(v=3-2t\) m s\(^{-1}\). Is its speed increasing or decreasing at \(t=2\)?
Answer
\(v(2)=-1\) and \(a=-2\). Same sign, so the speed is increasing.