Setting up. We choose the axis s oriented in the direction of motion, with the origin of times and positions fixed by the initial data. The motion consists of 3 stages and 4 events (A, B, C, D):
- Stage A→B: uniformly accelerated motion (a=3m/s2) for 4s;
- Stage B→C: uniform motion (a=0) for 100m;
- Stage C→D: braking to a stop over 500m.
It is convenient to organise the data in an events table, filling it in stage by stage:
| Event | t (s) | s (m) | v (m/s) | a (m/s²) |
|---|
| A | 4 | 2000 | 6 | |
| | | | 3 |
| B | 8 | 2048 | 18 | |
| | | | 0 |
| C | 13,56 | 2148 | 18 | |
| | | | −0,324 |
| D | 69,1 | 2648 | 0 | |
Stage A→B (acceleration). The final velocity and the final position are:
vBsB=vA+a(tB−tA)=6+3⋅4=18m/s=sA+vA⋅4+21⋅3⋅42=2000+24+24=2048m
Stage B→C (constant velocity, a=0). The time to cover the 100m is:
tC=tB+vBΔs=8+18100≈13,56ssC=2048+100=2148m
Stage C→D (braking, vD=0, Δs=500m). We obtain the acceleration from the time-independent relation:
vD2−vC2=2aΔs⇒0−182=2a⋅500⇒a=−1000324=−0,324m/s2
Finally, the stopping time and the final position:
tD=tC+avD−vC=13,56+−0,3240−18≈69,1ssD=2148+500=2648m
tD≈69,1s,sD=2648m,vD=0