Question

a farmer moves along boundary square field of side 10m in 40s. what will be magnitude of displacement of farmer at end of 2 min 20 sec

Answer 1

2min 20sec=140sec
oneside 40 sec
4sides 160 sec
___________________
20sec from completion
1/2 side
5m

Answer 2

Hey.

Here is your answer.

Given : Square field of side , a = 10 m

time taken to move along the boundary = 40s

Perimeter of the square field (boundary)

= 4× side

= 4 × 10

= 40 m

So, he travels 40 m in 40 s.

Now as time given : 3 minutes = 3 × 60 s

t = 180 s

Since 40s he takes to travel 40 m

so, in 180 s he will travel a distance of

$= \: \: \frac{40}{40} \times 180$

= 180 m

So, he will travel a total distance of 180m

along the side of the square

so, as the average speed

$= \: \frac{total \: distance}{total \: time}$

so, average speed of the person for 3 minutes

$= \: \frac{180}{180} \: \: \frac{m}{s} \\ \\ \\ = \: \: 1m {s}^{ - 1}$

Now , in 180 m he will take

$\frac{180}{40} \: rounds \: along \: the \: square \\ \\ i.e. \: 4 \: complete \: round \: and \: 20 \: m \: \\ from \: start$

So, it will be diagonally opposite from the start

as the sides are 10+10 m

So, the displacement = diagonal of the square

$\\ dispacement \: = \: side\sqrt{2} \\ \\ = \: \: side \sqrt{2} \: m$

As average velocity

$= \: \frac{displacement}{time} \\ \\ so \: avg \: velocity = \: \frac{10 \times \sqrt{2} }{180} \\ \\ = \: \frac{ \sqrt{2} }{18} \: m {s}^{ - 1}$

Hence, average speed is 1 m/s

whereas average velocity is 0.07856 m/s along the diagonal.

Thanks