Question

A boy travels 5km towards north,then he turns to right and travels another 5km before coming to rest.Calculate,(i)total distance travelled,(ii)total displacement

Answer 1

displacement -

$\sqrt{5 {}^{2} + 5 {}^{2} } = 5 \sqrt{2} cm$

distance-

$5 + 5 = 10cm$

Answer 2

Answer:

Displacement = 5√2 m N - E

Distance covered = 10 meter

Step-by-step explanation:

given that,

A boy travels 5km towards north,

then he turns to right

and travels another 5km

before coming to rest.

here,

let S1^-> = 5 km north

S2^-> = right of north = east

so,

for the resultant vector

θ = 90°

and we know that,

sum of vector

S^-> + S^-> = √(S² + S² + 2(S)(S) cosθ)

putting the values,

= √(5² + 5² + 2(5)(5) cos90°)

= √(25 + 25 + 50(0))

= √50

= 5√2 m

now,

since vectors are in north and east

then resultant displacement will be

in North -- East

so,

Displacement = 5√2 m N - E

now,

Distance covered = total path length

= 5 + 5 = 10 m

so,

________________

Displacement = 5√2 m N - E

Distance covered = 10 meter