Motion問題 (物理)

Firstly, calculate the distance travelled at the first 5 seconds :
t = 5 s, vi = 0, a = ?, d = ?
d = vit + 0.5at²
d = 0 + 0.5a(5)²
d = 12.5a m
Distance travelled at the first 5 second = 12.5a m

Secondly, calculate the distance travelled at the first 6 seconds :
t = 6 s, vi = 0, a = ?, d = ?
d = vit + 0.5at²
d = 0 + 0.5a(6)²
d = 18a m
Distance travelled at the first 6 second = 18a m

Distance travelled between the 5th and 6th seconds = Distance travelled duringthe 6th second
18a - 12.5a = 22
5.5a = 22
a = 4 m/s²
acceleration of the sports car = 4m/s²

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Motion問題 (物理)Calculate the uniform acceleration of a sports car which starts from rest and travels a distance of 22m during the sixth second of its motion.
請把過程列出
Equations of Motion:

1. Vf = Vi+at => Vf=0+6a=6*11/9=22/3 mps

2. d = Vit+0.5at^2 => 22=0+0.5*36a

Thus a=44/36=11/9 mps^2

3.Vf^2 = Vi^2+2ad => (22/3)^2=2*11/9*22

4.d = [(Vf+Vi)/2]t => 22=22/3*0.5*6

Vf = final velocity
Vi = initial velocity
d = displacement
a = acceleration
t = time
2012-03-05 14:18:28 補充 我算的答案大多是Vf=3.7m/s
可是正確答案是4ms/s.....可能有誤
到底怎麼算??