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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
avatar+27032 
+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(r2+s2) and angle = tan-1(s/r)

Alan  Jan 31, 2018
edited by Alan  Jan 31, 2018
 #2
avatar+89775 
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°

 

 

 

cool cool cool

CPhill  Jan 31, 2018

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