Question

A boy runs 100m along the east then he turns to his left side at right angles and run for 50m along the north direction find distance and magnitude of the displacement of the boy

Answer 1

Answer:

Distance = 150 m

Displacement = 50√5 m

Explanation:

given that,

A boy runs 100m along the east then he turns to his left side at right angles and run for 50m along the north

distance = total length covered by the boy

100 m + 50 m

= 150 m

now,

for displacement,

let 100 m along east be a^->

and

50 m along north be b^->

given angle = 90°

we know that,

in a vector addition,

a^-> + b^-> = a² + b² + 2ab cosθ

= √(100² + 50² + 2(100)(50) cos90°)

= √(10000 + 2500 + 10000 × 0)

= √(10000 + 2500)

= √12500

= 50√5 m

so,

displacement of the body

= 50√5 m

___________________

NOTE :

When the angle between the two vectors will 90° then sum of vectors =

√(a² + b²)

Answer 2

Answer:

Distance = 150 m

Displacement = 50√5 m

Explanation:

REFER TO ATTACHMENT

sorry for bad handwriting