Why displacement and distance are different numbers, and the integral that tells them apart.
A particle moves with velocity v = t² − 4t + 3 metres per second. The track is at the top, the velocity graph below it. Press play and watch the particle turn round twice.
Green area counts forwards. Red area counts backwards for displacement, and forwards for distance.
Differentiate to go one way, integrate to go back:
displacement s → velocity v = dsdt → acceleration a = dvdt = d²sdt²
Speed is the size of the velocity. A velocity of −2 m per second is a speed of 2 m per second. Negative speed does not exist, and writing it is a quick way to lose a mark.
The modulus is doing all the work. In practice you do not integrate a modulus directly: you find where v = 0, split the journey there, and add the sizes.
v = t² − 4t + 3 = (t − 1)(t − 3), so v = 0 at t = 1 and t = 3, and v is negative between them.
| leg | 0 to 1 | 1 to 3 | 3 to 4 | total |
|---|---|---|---|---|
| ∫v dt | ⅔ | −⅔ | ⅔ | ⅔ |
| distance | ⅔ | ⅔ | ⅔ | 4 |
Each leg happens to be the same size here, which makes the contrast as clean as it gets: the particle goes out, comes all the way back, then goes out again.
Finding where v = 0 is the first step, every time. Not because the question asks for it, but because without it you cannot split the integral, and without splitting it you cannot get the distance.
a = dv/dt = 2t − 4, which is zero at t = 2. That is where the particle is moving backwards fastest, not where it is stationary. Zero acceleration and zero velocity are completely different events, and questions exploit that.
1. At what times is the particle instantaneously at rest? Give the smaller one.
2. Over 0 ≤ t ≤ 4, which statement is right?
3. Find the acceleration at t = 0.
Solving v = 0 and splitting the integral there is the method mark on any total distance question. An unsplit integral gives the displacement and scores nothing for distance.
Read the word. Displacement, distance, speed and velocity are four different quantities and the question always names one precisely.
Your calculator will integrate the modulus directly if you ask it to, and that is a fine check, but show the split or the method marks are gone.
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