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.)

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

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

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s