Question

A body travels 4km towards north then turn to right and travels another 4km before coming to rest . Calculate the total distance travelled and totql displacement

Answer 1

Answer:

Given:

Body travels 4 km towards North and then turns right and travels another 4 km. Then comes to rest.

To find:

• Total Distance
• Total Displacement

Concept:

Distance is the total path length traversed by a body. So distance is always positive. It's a scalar Quantity.

Displacement is the shortest length between the starting and stopping point. It's a vector quantity representing both magnitude and direction.

So , displacement can be negative, zero or positive.

Calculation:

Total distance travelled :

$s = 4 + 4 = 8 \: km$

Total displacement can be calculated by Pythagoras theorem :

$d = \sqrt{ {4}^{2} + {4}^{2} } = 4 \sqrt{2} \: km$

Answer 2

Explanation:

$\bold{Given -} \\ A \: Body \: travel \: \bold{4 \: km} \: towards \: \bold{North} \\ then \: turn \: right \: and \: travel \: another \\ \bold{4 \: km} \: before \: coming \: to \: rest.$

$\bold{Total \: Distance \: travelled = } \\( 4 + 4) \: km \\ \bold{= 8 \: km} \\ \\ We \: know \: that, \\ \bold{Displacement -} \\ Shortest \: Distance \: from \: initial \: position \: \\ to \: final \: position. \\ \\ \bold{By \: Pythagoras \: theorem,} \\ (Hypotenuse) {}^{2} = (Perpendicular) {}^{2} + (Base) {}^{2} \\ \\ It \: means, \\ (BC) {}^{2} = {(AB)}^{2} + (AC) {}^{2} </p><p> \\ (BC) {}^{2} = (4) {}^{2} + (4) {}^{2} \\ BC = \sqrt{16 + 16} \\ BC = \sqrt{32} \\ \fbox\bold{BC = 4 \sqrt{2} \: km} \\ \\ Hence, \\ \fbox\bold{Displacement = 4 \sqrt{2} \: km}$