Question

A particle moves 3m towards north then 4m east and then finally 6m south.calculate the total distance and displacement

Answer 1

Answer:

5m

Explanation:

in AEF

EF=6-3=3m

AE=4m

using pythagoras theorem

AF2=AE2+EF2

=9+16

=25

TAKE UNDERROOT

$\sqrt{25 }$

5 m

Answer 2

Answer:

13m    (distance)

5m      (displacement)

Explanation:

Let north, south east be represented by N, S, E.

According to question:

⇒ 3m in North = 3N,  then

⇒ 4m in East(right) = 3N + 4E , then

⇒ 6m in South = 3N + 4E + 6S

Total displacement = 3N + 4E + 6S

But as we know, north is opposite to south, N = - S.

∴ displacement = 3(-S) + 4E + 6S = 4E + 3E  

As south and east are at right angle to each other, we can't them directly.

Using Pythagoras theorem,

Displacement = √3² + 4² = 5

Displacement is 5 m

But in terms of distance, we can add it directly as it doesn't depend on direction, distance = 3 + 4 + 6 = 13 m