At one instant a bicyclist is 40.0 m due east of a park's flag- pole, going due south with a speed of 10.0 m/s. then 30.0 s later, the cyclist is 40.0 m due north of the flagpole, going due east with a speed of 10.0 m/s. for the cyclist in this 30.0 s interval, what are the (a) magnitude and (b) direction of the displacement, the (c) magni- tude and (d) direction of the average velocity, and the (e) magni- tude and (f) direction of the average acceleration?

consider the east-west direction along the x-direction and north-south direction along the y-axis

a)

x₁ = initial position = 40 i + 0 j

x₂ = final position = 0 i + 40 j

displacement is given as

Δx = x₂ - x₁ = (0 i + 40 j) - (40 i + 0 j ) = - 40 i + 40 j

magnitude : sqrt((- 40)² + (40)²) = 56.6 m

b)

direction : 180 - tan⁻¹(40/40) = 135 deg counterclockwise from east

c)

average velocity is given as

average velocity = displacemen/time

average velocity = (- 40 i + 40 j )/30 = - 1.33 i + 1.33 j

magnitude : sqrt((- 1.33)² + (1.33)²) = 1.88 m/s

d)

direction : direction : 180 - tan⁻¹(1.33/1.33) = 135 deg counterclockwise from east

e)

v₁ = initial velocity = - 10 j

v₂ = final velocity = 10 i

average acceleration is given as

a = (v₂ - v₁)/t = (10 i - 10 j)/30 = 0.33 i - 0.33 j

magnitude : sqrt((0.33)² + (- 0.33)²) = 0.47 m/s

f)

direction : direction : 360 - tan⁻¹(0.33/0.33) = 315 counterclockwise from east

b would be the answer

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