A journey in three stages
Now let us use everything so far on one problem. A real journey usually has several stages.
The problem
An object starting from rest undergoes a uniform acceleration for 4 seconds and reaches a velocity of 12 m s−1. It then travels at that uniform velocity of 12 m s−1 for another 4 seconds. Finally it decelerates uniformly for 2 seconds and comes to rest.
Stage 1 — accelerating (0 – 4 s)
acceleration = (12 − 0) m s−1 ÷ 4 s = 3 m s−2
average velocity = (0 + 12) ÷ 2 = 6 m s−1
displacement = 6 m s−1 × 4 s = 24 m
Stage 2 — uniform velocity (4 – 8 s)
The velocity is constant, so no average is needed.
displacement = 12 m s−1 × 4 s = 48 m
Stage 3 — decelerating (8 – 10 s)
acceleration = (0 − 12) m s−1 ÷ 2 s = −6 m s−2
so the deceleration is 6 m s−2
average velocity = (12 + 0) ÷ 2 = 6 m s−1
displacement = 6 m s−1 × 2 s = 12 m
Total displacement
| Stage | Time | Displacement |
|---|---|---|
| Accelerating | 4 s | 24 m |
| Uniform velocity | 4 s | 48 m |
| Decelerating | 2 s | 12 m |
total displacement in 10 s = 24 + 48 + 12 = 84 m
So the object's final position is 84 m from its starting position in a straight line.
Remember the method
A multi-stage problem is solved by thinking about one stage at a time. Find each stage's displacement separately, then add them at the end. In the next section we learn to read this same journey off a graph.