Question

A farmer moves along the
boundary of a square field of side
10 m in 40 s. What will be the
magnitude of displacement of the
farmer at the end of 2 minutes 20
seconds from his initial position?​

Answer 1

Answer:

14.143

Explanation:

Total time=2min+20sec

=2×60+20

=120+20

=140sec

In 40sec number of rounds made =1

In 140sec number of rounds made =1/40×140

=3.5 rounds

${ac}^{2} = {ad }^{2} + {dc}^{2} \\ {ac}^{2} = {10}^{2} + {10 }^{2} \\ {ac }^{2} = 100 + 100 \\ ac = \sqrt{200} \\ ac = 14.143$

Answer 2

Answer:

the distance covered in one round of the field = the perimeter of the field

= 4×10m=40m

time taken = 40 m 40 second in one round

In 2 min 20 sec i.e. 140 sec the farmer cover a distance of 140m

140m = 140/ 40 = 3.5 round

0.5 × 40 =20 m

the displacement is AC

by Pythagoras theorem,

in right angle triangle ABC

AC² = AB²+ BC²

= (10)²+(10)²

= 100+100

= 200

AC = √200 = √100×2=10√2m

the magnitude of displacement of the farmer is 10√2 m.

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