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प्रश्न
In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?
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उत्तर १
Velocity of the boat, vb = 51 km/h
Velocity of the wind, vw = 72 km/h
The flag is fluttering in the north-east direction. It shows that the wind is blowing toward the north-east direction. When the ship begins sailing toward the north, the flag will move along the direction of the relative velocity (vwb) of the wind with respect to the boat
The angle between vw and (–vb) = 90° + 45°
`tanbeta = (51sin(90+45))/(72+51cos (90+45))`
`=(51sin45)/(72+51(-cos45)) = (51xx1/sqrt2)/(72-51xx1/sqrt2)`
`= 51/(72sqrt2-51) = 51/(72xx1.414 - 51) = 51/50.800`
`:.beta = tan^(-1) (1.0038) = 45.11^@`
Angle with respect to the east direction = 45.11° – 45° = 0.11°
Hence, the flag will flutter almost due east.
उत्तर २
When the boat is anchored in the harbour, the flag flutters along the N-E direction. It shows that the velocity of wind is along the north-east direction. When the boat starts moving, the flag will flutter along the direction of relative velocity of wind w.r.t. boat. Let Vwb be the relative velocity of wind w.r.t. boat and P be the angle between Vwb and vw (see fig. below)
Now `vecv_(wb) = vecv_w + (-vecv_b)`
Here `|vecv_w| = 72 km/h`
`|-vecv_b| = 51 km/h`
Angle between `vecv_w` and `-vecV_b` is 135^@. Then
`tan beta = (51 sin135^@)/(72+51cos135^@) = (51 sin 45^@)/(72+51(-cos 45^@))`
= `(51xx(1/sqrt2))/(72-51(1/sqrt2)) = 1.0039`
`:.beta = tan^(-1) (1.0039) = 45.1^@`
Angle w.r.t east direction = `45.1^@ - 45^@ = 0.1^@`
It means the flag will flutter almost due east.
उत्तर ३
When the boat is anchored in the harbour, the flag flutters along the N-E direction. It shows that the velocity of wind is along the north-east direction. When the boat starts moving, the flag will flutter along the direction of relative velocity of wind w.r.t. boat. Let Vwb be the relative velocity of wind w.r.t. boat and P be the angle between Vwb and vw (see fig. below)
Now `vecv_(wb) = vecv_w + (-vecv_b)`
Here `|vecv_w| = 72 km/h`
`|-vecv_b| = 51 km/h`
Angle between `vecv_w` and `-vecV_b` is 135^@. Then
`tan beta = (51 sin135^@)/(72+51cos135^@) = (51 sin 45^@)/(72+51(-cos 45^@))`
= `(51xx(1/sqrt2))/(72-51(1/sqrt2)) = 1.0039`
`:.beta = tan^(-1) (1.0039) = 45.1^@`
Angle w.r.t east direction = `45.1^@ - 45^@ = 0.1^@`
It means the flag will flutter almost due east.
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