Question

At one instant a bicyclist is 41.0 m due east of a park's flagpole, going due south with a speed of 15.0 m/s. Then 27.0 s later, the cyclist is 41.0 m due north of the flagpole, going due east with a speed of 15.0 m/s. For the cyclist in this 27.0 s interval, what are the

(a) magnitude and
(b) direction of the displacement, the
(c) magnitude and
(d) direction of the average velocity, and the
(e) magnitude and
(f) direction of the average acceleration? (Give all directions as positive angles relative to due east, where positive is meaured going counterclockwise.)

Answer 1

displacement is 41√2 m and 45º; velocity is 15 √2 m/s and 315º and the acceleration is 0.786 m/s² and 315º

Step-by-step explanation:

a and b) The cyclist is at a distance east and then north, we can find the displacement using the Pythagorean theorem

Xt² = X1² + X2²

Xt = √(41² + 41²)

Xt = 41 √2 m

The angle can be found by trigonometry

tan θ = X1 / X2

tan θ = 41/41 = 1

θ = 45º

c and d) the velocities are also vector magnitudes, so they can be brought in the same way as the displacements

V² = V1² + V2²

V = √ V1² + V2²

V = √[(-15)² + 15²

V = 15 √2 m/s

Tan θ = -15/15 = -1

θ = -45º

This is (270+ 45) measured from the east counterclockwise

θ = 315º

E and f) the average accelerations

Let's calculate the value of acceleration, which is also a vector

a = (Vf -Vo )/ t

The initial speed is directed south is negative and the final speed is directed east is positive

Vf = 15 m/s i^ θ = 0º

Vo = -15 m/s j^ θ = 270º

a = (15 i^ - (- 15) j^) / 27

cañculate the module

a² = (Vf² + Vo²) / t²

a = √ (15² + 15²) / 27

a = 15 √2 /27

a = 0.786 m/s²

The direction of acceleration is

tan θ = ay / ax

tan θ = -0.55 / 0.55 = -1

θ = -45º

This is (270+ 45) measured from the east counterclockwise