What are the magnitude Δ v and angle θ of the change in velocity if the initial velocity is 50.0 m/s south and the final velocity is 35.0 m/s west? Let the angle θ be measured relative to the east direction (e. g., θ = 90 ∘ means Δ → v points north, θ = 180 ∘ means Δ → v points west, etc.).

∆v = 61.0 m/s

θ = 305°

Explanation:

Given;

initial velocity u= 50.0 m/s angle 270°

Final velocity v = 35.0 m/s angle 180°

∆v = v - u 1

Resolving into x and y axis;

For y axis

∆vy = vsin180 - usin270

∆vy = 0 - (-50) = 50 m/s

x axis;

∆vx = vcos180 - ucos270

∆vx = -35 - 0= -35m/s

Magnitude of ∆v;

∆v = √(-35^2 + 50^2)

∆v = 61.0 m/s

Direction;

Tanθ = ∆vy/∆vx

θ = taninverse (50/-35)

θ = -55°

θ = 360-55 = 305°

graphing motion: mastery test

209be

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0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

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20

time (seconds)

answer: 0 meters/second^2