Question

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)​

Answer 1

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

Answer 2

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.......