Question

speed.
2. A man starts from a point and travels along a rectangular field of dimensions
500 m x 300 m. When he reaches the diagonally opposite end, find his distance
and displacement.​

Answer 1

Answer:

• Distance = 800 m
• Displacement = 583 m

Figure :

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Solution:

Distance is the actual distance travelled by a person . When the man reaches diagonally opposite end , he must have covered both length and breadth once i.e. if he started from B to reach diagonally opposite D , he covered BC and CD .

So distance travelled = BC + CD

= 500 + 300

= 800 m

Displacement is the difference between the initial and final point . Since , his final point is D and initial point is B , the displacement thus becomes BD .

From Pythagoras theorem ,

$\sf{hypotenuse}^{2} = {base}^{2} + {perpendicular}^{2}$

BD² = BC² + CD²

$\implies \tt BD = \sqrt{BC^2 + CD^2 }$

$\tt = \sqrt{ {500}^{2} + {300}^{2} } \\ \\ \tt = \sqrt{250000 + 90000} \\ \\ \tt = \sqrt{340000} \\ \\ \tt \: = 100 \times 5 . 83 \: m \\ \\ \tt = 583 \: m(approx)$

$\underline{\therefore \sf The\ distance\ is\ 800\ m\ and\ displacement\ is}\underline{\sf 583\ m}$

Answer 2

Answer:

Distance is the actual distance travelled by a person . When the man reaches diagonally opposite end , he must have covered both length and breadth once i.e. if he started from B to reach diagonally opposite D , he covered BC and CD .

So distance travelled = BC + CD

= 500 + 300

= 800 m

Displacement is the difference between the initial and final point . Since , his final point is D and initial point is B , the displacement thus becomes BD .

From Pythagoras theorem ,

\sf{hypotenuse}^{2} = {base}^{2} + {perpendicular}^{2}hypotenuse

2

=base

2

+perpendicular

2

BD² = BC² + CD²

\implies \tt BD = \sqrt{BC^2 + CD^2 }⟹BD=

BC

2

+CD

2

\begin{gathered} \tt = \sqrt{ {500}^{2} + {300}^{2} } \\ \\ \tt = \sqrt{250000 + 90000} \\ \\ \tt = \sqrt{340000} \\ \\ \tt \: = 100 \times 5 . 83 \: m \\ \\ \tt = 583 \: m(approx)\end{gathered}

=

500

2

+300

2

=

250000+90000

=

340000

=100×5.83m

=583m(approx)

\underline{\therefore \sf The\ distance\ is\ 800\ m\ and\ displacement\ is}\underline{\sf 583\ m}

∴The distance is 800 m and displacement is

583 m