Question

A farmer new around boundary to of a square field of side 10 m in 40 seconds what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his inital position farmer

Answer 1

★★Question;-

A farmer new around boundary to of a square field of side 10 m in 40 seconds what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his inital position farmer

★★Answer;-

$\boxed{\boxed{{=14.1m}}}$

★★Explanation;-

length of square filed = 10 m

Time taken to move along square filed = 40 sec

Total time the Farmer walked on square field =2min 20 sec

$=2×60+20$⠀⠀⠀ [ 1 min = 60 sec]

$= 120+20$

$= 140 sec$

In 40 sec Farmer covers 40 m

Thus,

Distance covered in 40 sec= 40m

Distance covered in 1 sec $=\frac{40}{40}m$

Distance covered in 140 sec $=\frac{40}{40}×140m$

$=140m$

Thus,

Total round $\boxed{\boxed{{=\frac{Total\: distance\: covered}{Distance\: covered\:in\:1\:round}}}}$

$=\frac{140}{40}$

$=3.5 round$

∴ The Farmer covered 3.5 rounds

If he started from A he will be at C after 20min 20 sec

Displacement = shortest distance

$=AC$

By Pythagoras theorem,

${Hypotenuse}^{2}={Height}^{2}+{Base}^{2}$

${AC}^{2}={AB}^{2}+{BC}^{2}$

${AC}^{2}={10}^{2}+{10}^{2}$

${AC}^{2}=100+100$

${AC}^{2}=200$

$AC={ \sqrt{200}}$

$AC={ \sqrt{2×100}}$

$AC=10{ \sqrt{2}}m$

$AC=10×1.41$

$\boxed{\boxed{{AC=14.1m}}}$