12. Two buses start from a bus stand with velocities
10 km h-1 and 30 km h- along two tracks incli-
ned at an angle 60°. Find the distance between
them after 12 minutes.
[Ans. 277 km)
Answers
Answered by
19
Answer:
2√ 7 km
Explanation:
Given :
- Speed of the first bus = 10 km/h
- Speed of the second bus = 30 km/h
- Time = 12 minutes = 1/5 hours
To find :
- Distance between them after 12 minutes at an inclination of 60°
Refer to the attachment for the distance to be found
So we can say BC = AC-AB in terms of vector
So using vector rule:
BC = √ ab²+ac²-2×ab×ac×(cos angle)
BC = √ 2²+6²-2×2×6×(cos 60°)
BC = √ 4 + 36 - 24 (cos 60°)
Cos 60° = ½
BC = √ 40-12
BC = √ 28
BC = 2 √ 7
The distance is equal to 2√7 km
Attachments:
Answered by
13
v₁ = 10 km/h
v₂ = 30 km/h
T = 12 Min = 1/5 h
AC = Distance Travelled with speed of 30 Kmph in 1/5 h = 30 × 1/5 = 6 Km
AB = Distance Travelled wifh speed of 10 Kmph in 1/5 h = 10 × 1/5 = 2 Km
Using Vector Rule,
|BC| = √A² + B² -2AB Cos60°
|BC| = √4+36 - 24/2
|BC| = √28
|BC| = √28 |BC| = 2√7 Km Ans.......
Attachments:
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