Question 6
Consider the same journey — accelerating from rest to 15 m s-1 in 6 s, 6 s at uniform velocity, then stopping in 3 s.
| t (s) | 0 | 6 | 12 | 15 |
|---|---|---|---|---|
| v (m s⁻¹) | 0 | 15 | 15 | 0 |
What is the object's displacement during the first 6 seconds?
The right answer is the one marked above.
45 m. The displacement over the first 6 s is the area of the triangle under that part of the graph:
displacement = ½ × 15 × 6 = 45 m
The average velocity route gives the same answer: (0 + 15) ÷ 2 = 7.5 m s-1, multiplied by 6 s gives 45 m.
90 m is the distance covered in the middle section at uniform velocity (15 × 6); 22.5 m is the distance covered in the last 3 s.