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A fighter plane traveling at u km/hr wants to refuel itself in mid-air. It sights a refueling station coming towards it with a speed of v km/hr. The angle of elevation from the refuelling station and the shortest distance between them are 30° and 220 km respectively. The plane sends a signal for refueling to the refueling station. It receives a confirmation in 10 minutes. During this time, the angle of elevation changes to 60°. If the vertical distance between them remains the same throughout and u : v = 10 : 1 , find the speed of the refueling station.
- a)40 km/hr
- b)40√3 km/hr
- c)45 km/hr
- d)45√3 km/hr
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A fighter plane traveling at u km/hr wants to refuel itself in mid-air...
The positions of the plane and the refueling station during the 10 minute span can be shown as follows,
The vertical distance (SQ or S'R) between the plane and the refueling station:
SQ = S'R = PS sin 30° = 220 sin 30° = 110 km
The initial horizontal distance (PQ) between the plane and the refueling station:
PQ = PS sin 30° = 220 sin 30° = 110√3 km
PQ = PS sin 30° = 220 sin 30° = 110√3 km
The final horizontal distance (P'R) between the plane and the refueling station:
P'R = S'R / tan 60° = 110 / tan 60° = 110 / √3 km
The distance travelled by the plane (PP') in 10 minutes = u / 6 km
The distance travelled by the refueling station (SS') in 10 minutes = v / 6 km = RQ
P'R = S'R / tan 60° = 110 / tan 60° = 110 / √3 km
The distance travelled by the plane (PP') in 10 minutes = u / 6 km
The distance travelled by the refueling station (SS') in 10 minutes = v / 6 km = RQ
Now, PQ = PP' + P'R + SS'
∴ 110√3 = 110 / √3 + u / 6 + v / 6 ...(i)
∴ 110√3 = 110 / √3 + u / 6 + v / 6 ...(i)
Also,
u / v = 10 / 1
u / v = 10 / 1
u = 10v
Substituting the value of u in (i),
∴ 110√3 = 110 / √3 + 11v / 6
∴ 110√3 = 110 / √3 + 11v / 6
Solving the given equation, we get,
v =40√3 km/hr
v =40√3 km/hr
Hence, option 2.
Most Upvoted Answer
A fighter plane traveling at u km/hr wants to refuel itself in mid-air...
Let's assume that the fighter plane is at point A and the refueling station is at point B. The shortest distance between the fighter plane and the refueling station is represented by line segment AB.
Given:
- The speed of the fighter plane is u km/hr.
- The speed of the refueling station is v km/hr.
- The angle of elevation from the refueling station to the shortest distance between them is 30 degrees.
We can represent the situation using a triangle, with the fighter plane, refueling station, and the shortest distance between them.
Using trigonometry, we can relate the angle of elevation, the shortest distance, and the distances traveled by the fighter plane and refueling station.
In triangle AOB:
- Angle AOB = 30 degrees (angle of elevation)
- Distance OA = u km/hr (distance traveled by the fighter plane)
- Distance OB = v km/hr (distance traveled by the refueling station)
We can use the trigonometric ratio of tangent to find the relationship between the angle of elevation and the shortest distance between the fighter plane and the refueling station.
tan(angle AOB) = shortest distance (AB) / distance traveled by the fighter plane (OA)
tan(30 degrees) = AB / u
Rearranging the equation, we get:
AB = u * tan(30 degrees)
Therefore, the shortest distance between the fighter plane and the refueling station is AB = u * tan(30 degrees).
Note: The units of speed (km/hr) cancel out in the equation, leaving us with the shortest distance in kilometers.
Given:
- The speed of the fighter plane is u km/hr.
- The speed of the refueling station is v km/hr.
- The angle of elevation from the refueling station to the shortest distance between them is 30 degrees.
We can represent the situation using a triangle, with the fighter plane, refueling station, and the shortest distance between them.
Using trigonometry, we can relate the angle of elevation, the shortest distance, and the distances traveled by the fighter plane and refueling station.
In triangle AOB:
- Angle AOB = 30 degrees (angle of elevation)
- Distance OA = u km/hr (distance traveled by the fighter plane)
- Distance OB = v km/hr (distance traveled by the refueling station)
We can use the trigonometric ratio of tangent to find the relationship between the angle of elevation and the shortest distance between the fighter plane and the refueling station.
tan(angle AOB) = shortest distance (AB) / distance traveled by the fighter plane (OA)
tan(30 degrees) = AB / u
Rearranging the equation, we get:
AB = u * tan(30 degrees)
Therefore, the shortest distance between the fighter plane and the refueling station is AB = u * tan(30 degrees).
Note: The units of speed (km/hr) cancel out in the equation, leaving us with the shortest distance in kilometers.
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A fighter plane traveling at u km/hr wants to refuel itself in mid-air. It sights a refueling station coming towards it with a speed of v km/hr. The angle of elevation from the refuelling station and the shortest distance between them are 30° and 220 km respectively. The plane sends a signal for refueling to the refueling station. It receives a confirmation in 10 minutes. During this time, the angle of elevation changes to 60°. If the vertical distance between them remains the same throughout and u : v = 10 : 1 , find the speed of the refueling station.a)40 km/hrb)40√3 km/hrc)45 km/hrd)45√3 km/hrCorrect answer is option 'B'. Can you explain this answer?
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