Displacement, velocity and acceleration, and the integral that tells distance apart from displacement.
A particle moves with velocity v = 3t² − 12t + 9 metres per second. The track is at the top and the velocity graph below. Press play and watch it turn round twice.
Green area counts forwards. Red counts backwards for displacement and forwards for distance.
s → v = dsdt → a = dvdt = d²sdt²
Speed is the magnitude of velocity. A velocity of −3 is a speed of 3. Negative speed does not exist.
You never integrate the modulus directly. Find where v = 0, split there, and add the sizes.
v = 3t² − 12t + 9 = 3(t − 1)(t − 3), so v = 0 at t = 1 and t = 3, and v is negative between them. Integrating, s = t³ − 6t² + 9t.
| t | 0 | 1 | 3 | 4 |
|---|---|---|---|---|
| s | 0 | 4 | 0 | 4 |
| leg | +4 | −4 | +4 |
Displacement over 0 to 4 is 4 − 4 + 4 = 4 m. Distance is 4 + 4 + 4 = 12 m, three times as far. The particle goes out, comes all the way back to its starting point, and goes out again.
Acceleration is not the same as turning. a = 6t − 12 is zero at t = 2, where the particle is moving backwards fastest. At t = 1 and t = 3 the velocity is zero but the acceleration certainly is not.
1. Find the displacement from t = 0 to t = 4.
2. Find the total distance travelled from t = 0 to t = 4.
3. At which time is the acceleration zero?
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 in the question. Displacement, distance, speed and velocity are four different quantities.
Units every time, and remember acceleration is in metres per second squared.
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