Respuesta :
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
[tex]R^2 = (4m/s)^2+(2m/s)^2 - 2(4m/s)(2m/s)cos(45)[/tex]
[tex]R = 2.95m/s[/tex]
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
[tex]d = 2.95(m/s)(10min)(\frac{60s}{1min})[/tex]
d = 1770m
The direction of the displacement is found using the law of sines with the velocity triangle
[tex]sin(\frac{\theta}{2}) = \frac{sin(45)}{2.9}[/tex]
Since the vehicle travels from south to north and from east to west then its deployment will be:
[tex]\theta = 29\°[/tex] north of due west
Answer:
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
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
