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The apparent path to intercept and average relative velocity?
I'm really lost in relative velocity. I do not know how to draw diagrams vector correctly, I feel so stupid. I do not seem the right direction, the angles and everything that needs to be known almoost to the relative velocity. Maybe I shall quote a simple QNS so we can start this step by step, so I can understand it:? here's a particular instant, two boats P and Q are 2 km away and P is north of Q. Ships move with constant speeds and direction shown: link to the diagram: 
(A) Speed and direction of the relationship Q P This question is asking about the speed and direction of motion of Q assuming P is "stationary" (in a relative sense). You need Draw a vector diagram and deduce the resulting vector is vq – vp. Oa-step by step procedure to draw the vector diagram is: (1) Draw the vector describing the motion ie Q 4m / s at 60 ° from the vertical, pointing approximately NW (2). Continue from the tip of the vector Q, draw the vector describing the motion of P in the opposite direction or That is, 3m / s to the east. (3) Draw the resultant vector connecting the start point of Q to the end point of P. If you draw the vector diagram to scale, you can measure the length of the resultant vector and its direction. Alternatively, you can also apply trigonometry to deduce the answers. (B) the distance between P and Q when Q East is the P. (1) Calculate the time required to move Q 2 km to the North. We are given that Q is moving at 4m / s to 60 ° west. Therefore, Q is moving at 2m / S to the north (ie 4m/sx cos60 °). The time required to move Q 2 miles north is 1000 (ie 2000 m / 2m). (2) Calculate the distance traveled west Q after 1000. 3464m distance traveled is (ie 4m/sx sin60 ° x 1000) (3). Calculate the distance P West after 1000. Distance traveled is 3000 (Ie 3m/sx 1000). (4) Calculate the distance between P and Q. Distance beyond = 3464 – 3000 = 464m
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