Question

A person travels 3 meter north , then 4 meter east and then climbs a tree of length 12m. Find the displacement​

Answer 1

distance covered is equal to the sum of three distances which is equal to = 3+4+12=19m

If initial position is considered to be at origin then the final position of the man is 3

i

^

+4

j

^

+12

k

^

So, the displacement covered by him =

3

2

+4

2

+12

2

=13m

Answer 2

Question:

A person travels 3 meter north , then 4 meter east and then climbs a tree of length 12m. Find the displacement.

Solution:

Let's assume that person reached a point D after traversing 3m North and 4m East. From the point D, the person climbs a tree.

(see the attachment)

$\sf OD^2= ON^2+DN^2 \\\\ \sf OD^2 =3^2+4^2 \implies 9+16 \implies 25 \\\\ \sf OD = \sqrt{25} \implies 5m.$

Now, from D, the person climbs a tree ( see the second attachment)

Let's denote the tree by DT

Now, calculate the displacement OT using the same above method.

$\sf OT^2= OD^2+DT^2 \\\\ \sf OT^2 = 5^2+12^2 \implies 25+144 \implies 169 \\\\ \sf OT = \sqrt{169} \implies 13m.$

Hence, the displacement is 13m.

Vector method:

Let the North direction be x-axis, East direction be y-axis and direction of the tree in z-axis. (see the 3rd attachment).

Then the displacement vector will be $\sf 3 \hat{i} + 4 \hat{j} + 12 \hat{k}$

Magnitude of displacement:

• $\sf \sqrt{3^2+4^2+12^2} \implies \sqrt{169} \implies 13m$