A total of 628 tickets were sold for the school play. They were either adult tickets or student tickets. There were 72 fewer stu
dent tickets sold than adult tickets. How many adult tickets were sold?
1 answer:
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Answer:
George is 43.20 ft East of his starting point.
Step-by-step explanation:
Let Paula's speed be x ft/s
George's speed = 9 ft/s
Note that speed = (distance)/(time)
Distance = (speed) × (time)
George takes 50 s to run a lap of the track at a speed of y ft/s
Meaning that the length of the circular track = y × 50 = 50y ft
George and Paula meet 14 seconds after the start of the run.
Distance covered by George in 14 seconds = 9 × 14 = 126 ft
Distance covered by Paula in 14 seconds = y × 14 = 14y ft
But the sum of the distance covered by both runners in the 14 s before they first meet each other is equal to the length of the circular track
That is,
126 + 14y = 50y
50y - 14y = 126
36y = 126
y = (126/36) = 3.5 ft/s.
Hence, Paula's speed = 3.5 ft/s
Length of the circular track = 50y = 50 × 3.5 = 175 ft
So, in 4 minutes (240 s), with George running at 9 ft/s, he would have ran a total distance of
9 × 240 = 2160 ft.
2160 ft around a circular track of length 175 ft, means that George would have ran a total number of laps (2160/175) = 12.343 laps.
Breaking this into 12 laps and 0.343 of a lap from the starting point. 0.343 of a lap = 0.343 × 175 = 60 ft
So, 60 ft along a circular track subtends an angle θ at the centre of the circle.
Length of an arc = (θ/360°) × 2πr
2πr = total length of the circular track = 175
r = (175/2π) = 27.85 ft
Length of an arc = (θ/360) × 2πr
60 = (θ/360°) × 175
(θ/360°) = (60/175) = 0.343
θ = 0.343 × 360° = 123.45°
The image of this incomplete lap is shown in the attached image,
The distance of George from his starting point along the centre of the circular track = (r + a)
But, a can be obtained using trigonometric relations.
Cos 56.55° = (a/r) = (a/27.85)
a = 27.85 cos 56.55° = 15.35 ft
r + a = 27.85 + 15.35 = 43.20 ft.
Hence, George is 43.20 ft East of his starting point.
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