✔ 最佳答案
(iii)
Let tₒ s be the time taken before timing started, and d m be the distance travelled before timing started.
When t = tₒ sec, s = d m:
s = ut + (1/2) at²
d = 0 + (1/2) × a × tₒ²
d = (1/2)atₒ²
When t = (tₒ + 4) sec, s = (d + 12) m:
s = ut + (1/2) at²
d + 12 = 0 + (1/2) × a × (tₒ + 4)²
atₒ² + 8atₒ + 16a = 2d + 24
Since d = (1/2)atₒ²: atₒ² + 8atₒ + 16a = at² + 24
a(tₒ + 2) = 3 …… [1]
When t = (tₒ + 4 + 5) sec = (tₒ + 9) sec and s = (d + 12 + 30) m = (d + 42) m:
s = ut + (1/2) at²
d + 42 = 0 + (1/2) × a × (tₒ + 9)²
atₒ² + 18atₒ + 81a = 2d + 84
Since d = (1/2)atₒ²: atₒ² + 18atₒ + 81a = atₒ² + 84
a(6tₒ + 27) = 28 …… [2]
[2]/[1]:
(6tₒ + 27) / (tₒ + 2) = 28/3
3(6tₒ + 27) = 28(tₒ + 2)
18tₒ + 81 = 28tₒ + 56
10tₒ = 25
tₒ = 2.5
Put tₒ = 2.5 into [1] :
a(2.5 + 2) = 3
a = 2/3
d = (1/2)atₒ² = (1/2) × (2/3) × 2.5² = 2.1
Distance travelled before timing started = 2.1 m
