Question

A particle is displaced 3m towards East, 4m towards North and then 5m vert
upwards. Find the displacement of the particle from the initial position
15m 2 3/2 m.
3 5√3m 4) zero​

Answer 1

Answer:

Given:

An object move 3 m towards East , 4 m towards North and finally 5 m vertically upwards.

To find:

Displacement of the particle

Concept:

Displacement is defined as the shortest path between the starting and stopping point of a motion.

It's a vector quantity having directions and magnitude.

Displacement can be negative , zero or positive.

Calculation:

As per the question, the particle has travelled in 3 Dimensions and the position vectors are located at 90° to one another .

So we can say :

$d = \sqrt{ {4}^{2} + {3}^{2} + {5}^{2} }$

$= > d = \sqrt{50}$

$= > d = 5 \sqrt{2} \: m$

So final answer is :

$\boxed{ \red{ \huge{ \bold{ d = 5 \sqrt{2} \: m}}}}$

Answer 2

SoLuTiOn :

➡ Given :

✏ A particle is displaced 3m towards east, 4m towards North and then 5m vertically upwards.

➡ To Find:

✏ The resultant displacement from its initial position is...

➡ Concept:

• Displacement is defined as the shortest distance between two points.
• Displacement can have zero magnitude.

➡ Calculation:

✏ The situation is as shown in the attachment...

$\rightarrow \rm \: oc = \sqrt{ {(oa)}^{2} + {(ab)}^{2} + ({bc)}^{2} } \\ \\ \rightarrow \rm \:oc = \sqrt{ {3}^{2} + {4}^{2} + {5}^{2} } \\ \\ \rightarrow \rm \: oc = \sqrt{9 + 16 + 25} \\ \\ \rightarrow \rm \: oc = \sqrt{2(25)} \\ \\ \rightarrow \: \underline{ \boxed{ \bold{ \rm{ \pink{displacement = 5 \sqrt{2} \: m}}}}} \: \star$

➡ Additional information:

• The magnitude of the displacement for a motion between two points may be zero but the corresponding path length is not zero.
• Displacement is vector quantity whereas path length is scalar quantity.
• Relation between displacement and path length is given by

$\star \: \underline{\boxed{ \bold{ \mathfrak{ \purple{path \: length \geqslant displacement}}}}} \: \star$