Question

4. Consider the three displacement vectors shown in the figure: Vector A has a magnitude of 8.10 km and a direction that makes an angle θ 290 to the left of the positive y-axis, vector B has a magnitude of S S0 km and a direction that makes an angle of a -25.0° above the positive x-axis, and vector C has a magnitude of 450 km and a direction that makes an angle β 600° below the negative x-axis. Determine the magnitude of the vector D A + E + km

Answer 1

net horizontal distance

Dx = - 8.1 sin 29 + 5.5 cos 25 + 4.5 cos 60

Dx = 3.308 km

net vertical distance

Dy = 8.1 cos 29 + 5.5 sin 25 - 4.5 sin 60

Dy = 5.537 km

net displacement

D^2 = Dx^2 + Dy^2

D = 6.428 km

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5.

during acceleration

d1 = 0.5 at^2

d1 = 0.5* 1.85* 11^2 = 111.925 m

velocity before deceleration, v = at = 1.85*11 = 20.35 m/s

distance travelled in deceleration

d2 = v^2/ ( 2 a')

d2 = 20.35^2 / ( 2* 3)

d2 = 69.02 m

net displacement, d = d1 + d2 = 180.945 m

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