1The idea in one paragraph

A particle moves along a straight line. Its displacement s is its position measured from a fixed origin O, with a sign that says which side of O it is on. Its velocity v is the rate of change of displacement, v = ds/dt, and its sign says which way it is moving. Its acceleration a is the rate of change of velocity, a = dv/dt = d²s/dt². Differentiating takes you down that chain, from s to v to a. Integrating takes you back up, as long as you know a starting value to fix each constant. Two quantities are easy to confuse and this page keeps them apart: displacement is where you end up relative to where you started, and it can cancel when you turn back; distance travelled is how much ground you cover, and it never cancels. The first is ∫v dt, the second ∫|v| dt.

2Displacement, velocity and acceleration

Picture a bead on a straight wire, or a lift in a shaft. Choose a fixed point O on the line and a positive direction, usually to the right or upwards.

• Displacement s(t): the signed position at time t. s > 0 means the particle is on the positive side of O, s < 0 the negative side, s = 0 at O. Units: metres, m.
• Velocity v(t) = ds/dt: how fast the displacement is changing. v > 0 means moving in the positive direction, v < 0 the negative direction, and v = 0 means at rest for that instant. Units: m s⁻¹.
• Acceleration a(t) = dv/dt = d²s/dt²: how fast the velocity is changing. Units: m s⁻².

Speed is the magnitude of velocity, |v|: how fast, without the direction. A velocity of −5 m s⁻¹ is a speed of 5 m s⁻¹. Speed is never negative.

Figure 1 is the whole of this page in one picture. Every question is a trip up or down that ladder.

3Going down the ladder: differentiate

A particle moves along a line so that its displacement from O after t seconds is s = t³ − 9t² + 24t metres, for t ≥ 0.

Now read the motion off these three functions.

Where does it start, and how fast? s(0) = 0, so it starts at O. v(0) = 24, so it sets off at 24 m s⁻¹ in the positive direction.

When is it at rest? At rest means v = 0: t = 2 and t = 4. Its positions then are s(2) = 8 − 36 + 48 = 20 m and s(4) = 64 − 144 + 96 = 16 m.

When does it change direction? Where v changes sign. v is positive for t < 2, negative for 2 < t < 4, and positive for t > 4. So it moves right, stops at 20 m and turns back, stops at 16 m and turns forward again. Figure 2 shows s against t; the particle is at rest where the graph is flat. Figure 3 shows the same motion as a path along the line itself, which is what actually happens.

A particle can be at rest and still accelerating. At t = 2 the velocity is 0 but a(2) = −6: the particle is stopped for an instant and already gathering speed in the negative direction, like a ball at the top of its throw.

Does it return to O? Only if s = 0 again: t(t² − 9t + 24) = 0. The quadratic has discriminant 81 − 96 < 0, so t = 0 is the only solution. It never comes back.

When is it moving backwards fastest? a = 0 at t = 3, which is where v has its minimum: v(3) = 27 − 54 + 24 = −3. So between t = 2 and t = 4 its greatest speed is 3 m s⁻¹, at t = 3. (Over 0 ≤ t ≤ 5, though, the greatest speed is 24 m s⁻¹ at t = 0, an end point, as 5.8 warned.)

4Speed, and speeding up or slowing down

A negative acceleration does not mean slowing down. It means the velocity is decreasing, and a velocity that is already negative and is decreasing becomes more negative: the particle goes faster. The rule is about signs.

A particle is speeding up when v and a have the same sign, and slowing down when they have opposite signs.

For the particle above, v = 3(t − 2)(t − 4) and a = 6t − 18. Figure 4 shades the four intervals.

Look at 2 < t < 3: the acceleration is negative and the particle is speeding up, because it is moving in the negative direction. That is exactly the case students get wrong.

To find a maximum speed over an interval, look for the largest value of |v|. That can be at a stationary point of v (where a = 0) or at an end of the interval, just as in 5.8. Check both.

5Going up the ladder: integrate

If you know the acceleration, integrate to get the velocity, then integrate again to get the displacement. Each integration brings a constant, and each constant needs a fact about a particular moment, usually t = 0. The words of the question carry those facts.

A particle has acceleration a = 6t − 12 m s⁻². Its initial velocity is 9 m s⁻¹, and at t = 0 it is 2 m to the left of O.

It is at rest at t = 1, where s(1) = 1 − 6 + 9 − 2 = 2, and at t = 3, where s(3) = 27 − 54 + 27 − 2 = −2. Find each constant before you integrate again; carrying an unknown C into the second integral turns it into Ct and doubles the algebra.

Never assume a constant is zero. If the question does not say where the particle starts, s(t) cannot be found, only the change in s between two times. That is the next section.

6Displacement and total distance

Integrating v from t₁ to t₂ gives the change in s, because s is an antiderivative of v:

Displacement from t₁ to t₂ = ∫ v(t) dt, from t₁ to t₂. Distance travelled from t₁ to t₂ = ∫ |v(t)| dt, from t₁ to t₂.

Both are in the formula booklet. The first counts motion in the negative direction as negative, so turning back cancels progress. The second takes the absolute value first, so every metre counts.

On a velocity–time graph these are areas, as Figure 5 shows for our particle from t = 0 to t = 5. The region above the axis is positive displacement, the region below is negative. Displacement adds them with their signs; distance adds their sizes.

On Paper 1 you cannot integrate |v| directly, so split the interval wherever v changes sign and add the sizes of the pieces. For v = 3t² − 18t + 24 on 0 ≤ t ≤ 5, v = 0 at t = 2 and t = 4, and an antiderivative is s(t) = t³ − 9t² + 24t.

Figure 3 shows why: 20 m out, 4 m back, 4 m out again. The particle ends 20 m from where it started but has covered 28 m of wire. If you integrate v straight from 0 to 5 you get 20 and have silently lost 8 m of travel: that is the commonest error on this page.

7On Paper 2: let the GDC do the integrals

When the velocity is awkward, the GDC finds the zeros, the maxima and the integrals, and your job is to set up each one correctly and write it down.

A particle moves with velocity v(t) = 3t e^−0.5t − 1 m s⁻¹ for 0 ≤ t ≤ 8. Figure 6 shows the graph.

When does it change direction? Use the zero tool on the graph of v: t = 0.409 s and t = 5.67 s. Both are sign changes, so both are changes of direction.

What is its greatest velocity? Use the maximum tool: v = 1.21 m s⁻¹ at t = 2. You can confirm by hand: a = 3e^−0.5t(1 − 0.5t), which is zero at t = 2. But the greatest speed could be at an end: |v(0)| = 1, less than 1.21, and |v(8)| = 0.560. So 1.21 m s⁻¹ is the greatest speed too.

What is its displacement over the 8 seconds?

What distance does it travel? Put the absolute value inside the integral on the GDC.

Distance is never less than the size of the displacement, and it equals it only if the particle never turns back. Here 4.73 > 2.90 because the particle moves backwards at the start and at the end. That is a good check on any answer.

If a question involves trigonometric velocities, check the GDC is in radians before you start. In degrees, sin t for t = 2 means the sine of 2°, and every answer will be wrong.