Question

A man walks 1.55 km south and then 2.25 km east, all in 2.80 hours.

(a)

What is the magnitude (in km) and direction (in degrees south of east) of his displacement during the given time?

magnitude km direction ° south of east

(b)

What is the magnitude (in km/h) and direction (in degrees south of east) of his average velocity during the given time?

magnitude km/h direction ° south of east

(c)

What was his average speed (in km/h) during the same time interval?

km/h

Answer 1

let's solve it using Cartesian plane

total time taken in journey = 2.8hours

initially man is at origin so it's position vector is 0i + 0j.

man walk 1.55 km in South so it's position vector is 0i - 1.55j

after this man walk 2.25km in East so it's n

position vector is 2.25i - 1.55j

a) displacement vector d = final position vector - initial position vector

so d = 2.25i - 1.55j - ( 0i + 0j)

d = 2.25i - 1.55j

magnitude of displacement |d| = √( (2.25)² + (-1.55)²)

so magnitude of displacement by man = 2.73 km

now direction ∅ = tan^​​​-1 (1.55/2.25)

so ∅ = 34.56°

so in direction 34.56° south of East

b) velocity vector v = change in position vector / total time taken

v = (2.25i - 1.55j - (0i - 0j))/2.8

so velocity vector = 0.8i - 0.55j

magnitude of velocity = √( (0.8)² + (-0.55)²)

magnitude of average velocity = 0.97km/h

direction of average velocity is always same as direction of displacement so

average velocity is in direction 34.56° south of East .

c) average speed equal = total distance traveled /total time

total distance traveled = 1.55 km in South + 2.25km in East

total distance traveled = 3.8 km

average speed equal = 3.8/2.8

so average speed equal to 1.36 m/s