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Question

A car is driven east for a distance of 50 km, then north for 30 km, and then in a direction 30° east of north for 25 km. Sketch the vector diagram and determine (a) the magnitude and (b) the angle of the car’s total displacement from its starting point.

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Solution

We label the displacement vectors A,B,and C ( and denote the result of their vector sum as r. we choose east as the ^i direction ( +x direction ) and north as the ^j direction ( +y direction . we note that the angle between C and the x axis is 600 . thus,
A=(50km)^i
B=(30km)^j
C=(25km)cos(600)^i+(25km)sin(600)^j

(a) The total displacement of the car from its initial position is represented by
r=A+B+C=(62.5km)^i+(51.7km)^j
which means that its magnitude is
|r|=(62.5km)2+(51.7km)2=81km

(b) The angle ( counterclockwise from +x axis ) is tan1(51.7km/62.5km)=400
which is to say that it point 400 north of east . Although the resultant r is shown in our sketch , it would be a direct line from the "tail "of A to he "head "of C

1632993_1764722_ans_a217dfc912234e3aab5b8c3ff3b6ed80.PNG

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