Question

A sailboat is being propelled westerly by the wind at a speed of 4 m/s. If the current is fl owing at 2 m/s to the northeast, where will the boat be in 10 min with respect to its starting position?

Answer 1

The resultant displacement is

R = √(-1551.47^2 + 848.53^2)

R = √3127062.3218

R = 1768.35m

The angle is given as

Tanx = 848.53/1551.47

x = taninverse(848.53/1551.47)

x = 28.68°

Therefore the sailboat is 1768.35m at angle 28.68° north west

Step-by-step explanation:

Using vector to resolve the problem

Where north represent +j (y plane) and east represent +i (x plane)

The velocity of the boat v(t) can be written as

The sailboat is being propelled westerly by the wind at a speed of 4 m/s which gives = -4i ( west is negative)

the current is fl owing at 2 m/s to the northeast,

= 2cos45i + 2sin45j

The total velocity is given as

v = -4i + 2cos45i + 2sin45j

v = (2cos45 -4)i + 2sin45j

Since,

Displacement d = velocity × time

Time = 10min = 600s

d = (2cos45-4)600i + (2sin45)600j

d = -1551.47i + 848.53j

The resultant displacement is

R = √(-1551.47^2 + 848.53^2)

R = √3127062.3218

R = 1768.35m

The angle is given as

Tanx = 848.53/1551.47

x = taninverse(848.53/1551.47)

x = 28.68°

Therefore the sailboat is 1768.35m at angle 28.68° north west

Answer 2

To solve this problem it will be necessary to apply the linear motion kinematic equations plus the law of cosines and sines. This in order to determine the resulting speed and direction of movement. Through the law of cosines we know that the final magnitude of the velocity would be

`R^2 = (4m/s)^2+(2m/s)^2 - 2(4m/s)(2m/s)cos(45)`

`R = 2.95m/s`

With the speed found now we can identify the distance traveled, because the displacement was made during a period of 10 minutes at the speed previously found

`d = 2.95(m/s)(10min)((60s)/(1min))`

d = 1770m

The direction of the displacement is found using the law of sines with the velocity triangle

`sin((\theta)/(2)) = (sin(45))/(2.9)`

Since the vehicle travels from south to north and from east to west then its deployment will be:

`\theta = 29\°` north of due west