Question

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?

Answer 1

Velocity of the boat, v_b = 51 km/h

Velocity of the wind, v_w = 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 (v_wb) of the wind with respect to the boat

The angle between v_w and (–v_b) = 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.

Answer 2

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 V_wb be the relative velocity of wind w.r.t. boat and P be the angle between V_wb 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.

Answer 3

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 V_wb be the relative velocity of wind w.r.t. boat and P be the angle between V_wb 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.