Question

At one instant a bicyclist is 55 m due east of a park's flagpole, going due south with a speed of 20 m/s. Then, 10 s later, the cyclist is 75 m due north of the flagpole, going due east with a speed of 20 m/s. For the cyclist in this 10 s interval, find each of the following?

Answer 1

Question

Find each of the following

(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?

Answer:

A. 93.01m

B.-53.75°

C. -5.5i + 7.5j m/s

D. 9.301m/s

E. 2.83 m/s²

F. 45.1°

Given

r0 = -55i m

v0 = -20i m/s

10 seconds later

r = 75j m

v = 20j m/s

Explanation:

A.

Magnitude of displacement = √(r0)² + (r)²

Magnitude = √-55² + 75²

Magnitude = √3025 + 5625

Magnitude = √8650

Magnitude = 93.00537618869137

Magnitude of displacement= 93.01 m -------------- Approximated

B.

Direction of displacement

TanФ = r/r0

Ф = tan-1 (r/r0)

Ф = tan-1 (75/-55)

Ф = tan-1 (-1.364)

Ф = −53.75°

C.

Magnitude of Velocity

Average velocity = Change in distance / Change in time

Average Velocity = (-55i + 75j)/10

Average Velocity = -5.5i + 7.5j m/s ------- Magnitude of velocity

D.

Direction of velocity = |Magnitude of displacement / time

Direction of velocity = 93.01/10

Direction of velocity = 9.301m/s

E.

Magnitude of Acceleration

Average Acceleration = Average Velocity/Time

Average Acceleration = Change in velocity/ Time

Average Acceleration = (v - v0)/t

= (20i - (-20j))10

Average Acceleration = 2i m/s² + 2j m/s²

Magnitude of Acceleration = √(2)² + (2)²

Magnitude = √4+4

Magnitude= √8

Magnitude of Acceleration = 2.828427124746190

Magnitude of Acceleration = 2.83 m/s² ------ Approximated

F.

Direction of Acceleration

Using ai = |a|cosФ

cosФ = |a|/ai

Ф = cos-1(|a|/ai)

Ф = cos-1(2/2.83)

Ф = cos-1(0.7061)

Ф = 45.08151997214929

Ф = 45.1° ---------- approximated