Hi there i a completely lost with this vectors question

A fruit bat has a speed in still air of 10ms−1. It is pointed in the direction of the bearing 130◦, but there is a wind blowing at a speed of 5ms−1 from the south-west.

Take unit vectors i to point east and j to point north.

(a) Express the velocity b of the bat relative to the air and the velocity w of the wind in component form, giving the numerical values in ms−1 to two decimal places. [7]

(b) Express the resultant velocity v of the bat in component form, giving numerical values in ms−1 to two decimal places. [3]

(c) Hence ﬁnd the magnitude and direction of the resultant velocity v of the bat, giving the magnitude in ms−1 to two decimal places and the direction as a bearing to the nearest degree.

DN83DN83 Jan 31, 2018

#1**+2 **

Here are some pointers to the right direction. I’ll leave you to fill in the details:

(a) b = 10*cos(130°)*i + 10*sin(130°)*j

w = 5*cos(45°)*i + 5*sin(45°)*j

(b) v = b + w. (add corresponding terms in i and j)

(c) If v = r*i + s*j then |v| = sqrt(r^{2}+s^{2}) and angle = tan^{-1}(s/r)

Alan Jan 31, 2018

#2**0 **

Let's convert "bearing" into standard degrees....we can convert back at the end.

A bearing of 130° is 320°

A wind * from* the SW means that the wind is blowing at a 45° angle

a) Velocity of the bat =

[ 10cos (320°) i, 10 sin (320°) j ] =

[ 7.66 i, - 6.43 j ]

Wind velocity = [ 5cos (45°) i , 5 sin (45°) j ] =

[ 3.54 i, 3.54 j ]

b) Velocity of the bat =

[ (7.66 + 3.54) i , ( -6.43 + 3.54 j ] =

[ 11.2 i ,- 2.89 j ]

c) Resultant velocity of the bat =

sqrt [ 11.2^2 + 2.89^2 ] =

11.57 m/s

Final direction = arctan (-2.89/ 11.2) = -14.47° = -14° [to the nearest degree ]

Converting this to bearing, we have 90 + 14 = 104°

CPhill Jan 31, 2018