A particle has velocity v = 6t − 3t2 m s−1 for 0 ≤ t ≤ 3. Find the total distance travelled.
- v = 3t(2 − t) = 0 at t = 0 and t = 2, and v changes sign at t = 2
- ∫v dt = 3t2 − t3. From 0 to 2: (12 − 8) − 0 = 4 m
- From 2 to 3: (27 − 27) − (12 − 8) = −4, so 4 m backwards
- Total distance = 4 + 4 = 8 m (the displacement is 0 m)
Answer: 8 m