As it passes over Grand Bahama Island, the eye of the hurricane is moving in a d

Question

As it passes over Grand Bahama Island, the eye of the hurricane is moving in a direction 60.0 degrees north of west with a speed of 41.0 km/h.
a) What is the unit-vector expression for the velocity of the hurricane?

It maintains this velocity for 3.00 h, at which time the course of the hurricane suddenly shifts due north, and its speed slows to a constant 25.0 km/h. This new velocity is maintained for 1.50 h.
b) What is the unit-vector expression for the new velocity of the hurricane?
c) What is the unit-vector expression for the displacement of the hurricane during the first 3.00h?
d) What is the unit-vector expression for the displacement of the hurricane during the latter 1.50 h?
e) How far from Grand Bahama is the eye 4.50 h after it passes over the island?

Explanation / Answer

   Let the unit vector along north be n^, west be w^.    a.   component of along north   (v1)n   =   v1 * sin ?                                                             =   41.0 * sin 600                                                             =   35.51   km/h          component of along west   (v1)w   =   v1 * cos ?                                                             =   41.0 * cos 600                                                             =   20.50   km/h    velocity vector in component form      v1   =   (v1)n n^   +   (v1)w w^                                                                   =   (35.51 n^   +   20.50 w^)   km/h

b.   After change of course,   v2   =   25.0   km/h, due north    hence angle with west    a  =   900    v2   =  v2 * sin a n^   +   v2 * cos a w^             =   25.0 * sin 900 n^   +   25.0 * cos 900 w^          =   25.0 n^      km/h    c.   displacement   r   =   v * t    =>   r1   =   v1 * t1                =   (35.51 n^   +   20.50 w^) * 3.00                =   (106.53 n^   +   61.50 w^)      km    d.   r2   =   25.0 n^ * 1.50                =   37.50 n^      km    e.   net displacement      r   =   r1   +   r2                                              =   106.53 n^   +   61.50 w^   +   37.50 n^                                              =   144.03 n^   +   61.50 w^    magnitude of displacement   |r|   =   v(|rn|2   +   |rw|2)                                                    =   v(144.032   +   61.502)                                                    =   156.61   km                                                                                                                 =   35.51   km/h          component of along west   (v1)w   =   v1 * cos ?                                                             =   41.0 * cos 600                                                             =   20.50   km/h    velocity vector in component form      v1   =   (v1)n n^   +   (v1)w w^                                                                   =   (35.51 n^   +   20.50 w^)   km/h

b.   After change of course,   v2   =   25.0   km/h, due north    hence angle with west    a  =   900    v2   =  v2 * sin a n^   +   v2 * cos a w^             =   25.0 * sin 900 n^   +   25.0 * cos 900 w^          =   25.0 n^      km/h    c.   displacement   r   =   v * t    =>   r1   =   v1 * t1                =   (35.51 n^   +   20.50 w^) * 3.00                =   (106.53 n^   +   61.50 w^)      km    d.   r2   =   25.0 n^ * 1.50                =   37.50 n^      km    e.   net displacement      r   =   r1   +   r2                                              =   106.53 n^   +   61.50 w^   +   37.50 n^                                              =   144.03 n^   +   61.50 w^    magnitude of displacement   |r|   =   v(|rn|2   +   |rw|2)                                                    =   v(144.032   +   61.502)                                                    =   156.61   km                                                                                                                 =   41.0 * cos 600                                                             =   20.50   km/h    velocity vector in component form      v1   =   (v1)n n^   +   (v1)w w^                                                                   =   (35.51 n^   +   20.50 w^)   km/h

b.   After change of course,   v2   =   25.0   km/h, due north    hence angle with west    a  =   900    v2   =  v2 * sin a n^   +   v2 * cos a w^             =   25.0 * sin 900 n^   +   25.0 * cos 900 w^          =   25.0 n^      km/h    c.   displacement   r   =   v * t    =>   r1   =   v1 * t1                =   (35.51 n^   +   20.50 w^) * 3.00                =   (106.53 n^   +   61.50 w^)      km    d.   r2   =   25.0 n^ * 1.50                =   37.50 n^      km    e.   net displacement      r   =   r1   +   r2                                              =   106.53 n^   +   61.50 w^   +   37.50 n^                                              =   144.03 n^   +   61.50 w^    magnitude of displacement   |r|   =   v(|rn|2   +   |rw|2)                                                    =   v(144.032   +   61.502)                                                    =   156.61   km                                                                                                  =   106.53 n^   +   61.50 w^   +   37.50 n^                                              =   144.03 n^   +   61.50 w^    magnitude of displacement   |r|   =   v(|rn|2   +   |rw|2)                                                    =   v(144.032   +   61.502)                                                    =   156.61   km                                                    

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