Ch03-p072 - 72 The ants trip consists of three displacements d1 =(0.40 m(cos 225 sin 225 =(0.28 m(0.28 m i j i j d =(0.50 m i 2 d3 =(0.60 m(cos 60

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(h) The y component of the net displacement net d G is ,123 ( 0.28 m) (0 m) (0.52 m) 0.24 m. net y y y y dd d d =++= + + = (i) The magnitude of the net displacement is 22 2 2 ,, (0.52 m) (0.24 m) 0.57 m. net net x net y d =+ = + = (j) The direction of the net displacement is , 11 , 0.24 m tan tan 25 (north of east) 0.52 m net y net x d d θ −− §· == = ° ¨¸ ©¹ If the ant has to return directly to the starting point, the displacement would be net d G . (k) The distance the ant has to travel is | | 0.57 m. net d −= G (l) The direction the ant has to travel is 25 (south of west) ° . 72. The ant’s trip consists of three displacements: 1 2 3 ˆˆ ˆ ˆ (0.40 m)(cos225 i sin 225 j) ( 0.28 m)i ( 0.28 m) j ˆ (0.50 m)i ˆ ˆ (0.60 m)(cos60 i sin 60 j) (0.30 m)i (0.52 m)j, d d d + ° = + = + ° = + G G G where the angle is measured with respect to the positive
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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