This note uses differentiation and integration to link displacement, velocity and acceleration for motion in a straight line. It shows how to find when an object is at rest, how far it travels, and how to tell distance apart from displacement. These ideas describe cars, lifts, projectiles and many other moving systems.
Key ideas
Take motion along a line, with time \(t\ge 0\). The displacement \(s(t)\) is the position relative to a fixed origin, and it can be positive or negative. The velocity \(v(t)\) is the rate at which the displacement changes, and the acceleration \(a(t)\) is the rate at which the velocity changes: \[v=\frac{ds}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2s}{dt^2}.\] Going the other way, integration recovers the earlier quantity: \(v=\int a\,dt\) and \(s=\int v\,dt\). Each integration brings a constant, which you find from given information such as \(s(0)=0\) or \(v(0)=3\).
Sign of the velocity. The sign shows the direction. If \(v\gt 0\), the object moves in the positive direction. If \(v\lt 0\), it moves in the negative direction. If \(v=0\), it is momentarily at rest, and this is often where it turns round. Speed is \(|v|\), which is never negative.
Speeding up or slowing down. The object is speeding up when \(v\) and \(a\) have the same sign, and slowing down when they have opposite signs. If \(a\) is positive and \(v\) is negative, the object is slowing down while moving in the negative direction.
Change in position and total distance. Over \(t_1\le t\le t_2\):
- The displacement (change in position) is \(\int_{t_1}^{t_2}v\,dt=s(t_2)-s(t_1)\). Forward and backward motion partly cancel.
- The total distance travelled is \(\int_{t_1}^{t_2}|v|\,dt\). Every part of the journey adds on, because speed is never negative.
To find the total distance by hand, solve \(v=0\), split the interval at every time the object turns, find the displacement over each part, and add the sizes. On the GDC, you can evaluate \(\int|v|\,dt\) directly by entering the absolute value function.
Maximum speed and greatest displacement. The greatest or least value of \(v\) occurs where \(a=0\) (check the ends of the interval too). The greatest or least value of \(s\) occurs where \(v=0\). When the motion is given by trigonometric functions, use radians.
Using graphs. On a velocity–time graph, the gradient is the acceleration and the signed area under the graph is the displacement. Area above the axis counts as positive and area below counts as negative. Sketching the graph on your GDC first often shows where the velocity changes sign.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(v=\dfrac{ds}{dt}\) | Velocity from displacement. |
| \(a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}\) | Acceleration from velocity or displacement. |
| \(s=\int v\,dt,\quad v=\int a\,dt\) | Reverse the process. Use given conditions to find each constant. |
| \(\int_{t_1}^{t_2}v\,dt\) | Displacement (change in position) from \(t_1\) to \(t_2\). |
| \(\int_{t_1}^{t_2}|v|\,dt\) | Total distance travelled from \(t_1\) to \(t_2\). |
| Speed \(=|v|\) | Size of the velocity, always non-negative. |
Worked examples
Example 1 A robot moves along a track so that its displacement from a start marker is \(s=t^3-6t^2+9t\) metres, where \(t\) is in seconds. For \(0\le t\le 4\), find when it is at rest, its acceleration at those times, and the total distance it travels.
\(v=\dfrac{ds}{dt}=3t^2-12t+9=3(t-1)(t-3)\). At rest when \(v=0\): \(t=1\) and \(t=3\).
\(a=\dfrac{dv}{dt}=6t-12\). At \(t=1\): \(a=-6\) m s\(^{-2}\). At \(t=3\): \(a=6\) m s\(^{-2}\).
Positions: \(s(0)=0\), \(s(1)=1-6+9=4\), \(s(3)=27-54+27=0\), \(s(4)=64-96+36=4\).
The robot goes forward 4 m (0 to 1 s), back 4 m (1 to 3 s), then forward 4 m (3 to 4 s).
Total distance \(=4+4+4=12\) m. The displacement is only \(s(4)-s(0)=4\) m.
Answer: at rest at 1 s and 3 s; accelerations \(-6\) and \(6\) m s\(^{-2}\); total distance 12 m
Example 2 A lift starts at the ground floor with \(s(0)=0\) and an upward velocity \(v(0)=3\) m s\(^{-1}\). Its acceleration is \(a=4-2t\) m s\(^{-2}\). Find \(v(t)\) and \(s(t)\), the greatest velocity, and the displacement at \(t=3\).
Integrate \(a\): \(v=4t-t^2+c\). Using \(v(0)=3\) gives \(c=3\), so \(v=3+4t-t^2\).
Integrate \(v\): \(s=3t+2t^2-\dfrac{t^3}{3}+k\). Using \(s(0)=0\) gives \(k=0\).
The velocity is greatest where \(a=0\): \(4-2t=0\), so \(t=2\). Then \(v=3+8-4=7\) m s\(^{-1}\). This is a maximum because \(a\) changes from positive to negative.
At \(t=3\): \(s=9+18-9=18\) m.
Answer: \(v=3+4t-t^2\), \(s=3t+2t^2-\dfrac{t^3}{3}\); greatest velocity 7 m s\(^{-1}\); \(s(3)=18\) m
Example 3 A piston moves with velocity \(v=3\cos(2t)\) m s\(^{-1}\), \(0\le t\le 2\) (radians). Find the displacement and the total distance travelled.
Set your GDC to radians. Displacement: \(\int_0^2 3\cos(2t)\,dt=\left[\dfrac32\sin 2t\right]_0^2=\dfrac32\sin4\approx -1.135\) m.
The velocity changes sign when \(\cos 2t=0\), which in this interval is at \(t=\dfrac\pi4\approx0.785\). (The next zero, \(t=\dfrac{3\pi}{4}\approx 2.36\), is outside the interval.)
For \(0\le t\le\dfrac\pi4\): \(v\ge 0\), and the distance is \(\left[\dfrac32\sin2t\right]_0^{\pi/4}=1.5\) m.
For \(\dfrac\pi4\le t\le 2\): \(v\le0\), and the distance is \(\left|\dfrac32\sin4-\dfrac32\right|=|-1.135-1.5|=2.635\) m.
Total distance \(=1.5+2.635=4.135\) m. A check with \(\int_0^2|3\cos 2t|\,dt\) on the GDC gives the same value.
Answer: displacement \(\approx -1.14\) m; total distance \(\approx 4.14\) m
Common errors
- Using \(\int v\,dt\) as the total distance. This gives displacement, and negative parts cancel positive ones.
- Forgetting to find the constant of integration when initial conditions are given, such as \(v(0)=3\).
- Saying the object is slowing down just because \(a\) is negative. Compare the signs of \(v\) and \(a\).
- Leaving the calculator in degree mode for trigonometric motion.
- Treating speed as negative. Speed is \(|v|\); only velocity has a direction.
Exam tip: Start any total-distance question by solving \(v=0\) in the given interval. List the turning times and the positions at each, then add the distances between them. State units for every answer: m, m s-1 or m s-2.
Check yourself
- A particle has displacement \(s=2t^3-9t^2+12t\). Find when it is at rest.
Answer
\(v=6t^2-18t+12=6(t-1)(t-2)=0\), so \(t=1\) and \(t=2\). - The velocity is \(v=2t+1\) and \(s(0)=3\). Find \(s(2)\).
Answer
\(s=t^2+t+c\), and \(s(0)=3\) gives \(c=3\). So \(s(2)=4+2+3=9\). - A cyclist's velocity is \(v=3-t\) for \(0\le t\le5\). Find the displacement and the total distance.
Answer
\(s=3t-\dfrac{t^2}2\) (taking \(s(0)=0\)). The velocity is zero at \(t=3\), where \(s=4.5\). At \(t=5\): \(s=15-12.5=2.5\). Displacement \(=2.5\). Distance \(=4.5+(4.5-2.5)=6.5\). - The acceleration is \(a=6t-4\) and \(v(0)=2\). Find the velocity at \(t=2\).
Answer
\(v=3t^2-4t+c\), and \(v(0)=2\) gives \(c=2\). So \(v(2)=12-8+2=6\).