A particle undergoes three successive displacements in a plane, as follows: 1, 1
ID: 1688443 • Letter: A
Question
A particle undergoes three successive displacements in a plane, as follows: 1, 1.24 m southwest; then 2, 4.73 m east; and finally 3, 7.74 m in a direction 42.6o north of east. Choose a coordinate system with the y axis pointing north and the x axis pointing east. What are (a) the x component and (b) the y component of 1? What are (c) the x component and (d) the y component of 2? What are (e) the x component and (f) the y component of 3?What are (g) the x component, (h) the y component, and (i) the magnitude of the particle's net displacement? (j) If the particle is to return directly to the starting point, how far should it move?
Explanation / Answer
given: a)d_1x = 1.24cos3150 = 1.24(0.70) = 0.87 m = 1.24(0.70) = 0.87 m b)d_1y = 1.2sin3150= 1.24 (-0.70) = -0.87 m c)d_2x = 2.47cos00 = 2.47 m d)d_2y = 2.47sin00 = 0 e)d_3x = 7.74(cos42.60) = 7.74(0.73) = 5.69 m f)d_3y = 7.74(sin42.60) = 7.74(0.67) = 5.23 m g)total displacement in xcomponent D_x = d_1x+d_2x+d_3x = 0.87+2.47+5.69 = 9.03 m h) net displacement in y component D_y = d_1y+d_2y+d_3y = -0.87+0+5.23 = 4.36 m i) net displacement D = vD_x2+D_y2 = v(9.03)2+(4.36)2 = v100.54 = 10.02 m j) when the particle returns to its original point directly it covers the distance 10.02 m j) when the particle returns to its original point directly it covers the distance 10.02 m
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