Question

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

please solve these questions quickly.

I want 1) , 2) and 3) answers please solve it
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Answer 1

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Given side of square =10m, thus perimeter P=40m

Time taken to cover the boundary of 40 m =40 s

Thus in 1 second, the farmer covers a distance of 1 m

Now distance covered by the farmer in 2 min 20 seconds = 1×140=140m

Now the total number of rotation the farmer makes to cover a distance of 140 meters =totaldistance/perimeter

=3.5

At this point, the farmer is at a point say B from the origin O

Thus the displacement s= √10²+√10²

from Pythagoras theorem.

s=10 √2 =14.14m

Answer 2

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Given that a farmer moves along the boundary of square field of side 10m in 40s.

We need find the magnitude of displacement at 2 minutes 20 seconds or 140 s from his initial position.

First, Let us find the speed of farmer.

⇒ Distance Travelled = Perimeter of the field

⇒ d = 4×Side

⇒ d = 40 m

Now,

⇒ Speed = Distance / Time

⇒ Speed = 40 / 40

⇒ Speed = 1 m/s ...(1)

Let us find the distance travelled by farmer at 140 s,

⇒ Distance = Speed × Time

⇒ Distance = 1 × 140

⇒ Distance = 140 m

Now, Let us find how many times he would go around the field (Given, Initial position is the first vertex of the square)

⇒ Number of rounds = Distance travelled / Perimeter of field

⇒ rounds = 140 / 40

⇒ rounds = 14/4

⇒ rounds = 3.5

At 3 rounds his displacement was 0 but he then travelled half the round.

Which means, The diagonal of the field is the displacement of farmer,

⇒ Displacement = √2 × Side

⇒ Displacement = √2 × 10

⇒ Displacement = 10√2 m

$\huge\mathcal{{\colorbox{navy}{{\colorbox{lightblue}{Answer \: 2 \: }}}}}$

Given, a bus decreases its speed from initial velocity u = 80 km/h to final velocity v = 60 km/h in time t = 5s.

We need to find the acceleration.

But, first let us convert velocities in m/s

u = 22.22 m/s

v = 16.67 m/s

We know, Acceleration is the change in velocity per unit time.

⇒ a = (v - u) / t

⇒ a = (22.22 - 16.67) / 5

⇒ a = 5.55 / 5

⇒ a = 1.11 m/s

Hence, Acceleration of the bus is 1.11 m/s.

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A train starting from rest (Initial velocity, u = 0 m/s)

attains a final velocity of v = 40 km/h in time t = 10 minutes

Convert velocities, time into m/s & s respectively.

u = 0 m/s

v = 11.11 m/s

t = 600 s

We know, Acceleration is the change in velocity per unit time.

⇒ a = (v - u)/t

⇒ a = (11.11 - 0)/600

⇒ a = 11.11 / 600

⇒ a = 0.0185 m/s

⇒ a = 1.85 × 10⁻² m/s

or, You can convert it into km/hr

⇒ a = 0.0185 × 3600 / 1000

⇒ a = 0.0185 × 3.6

⇒ a = 0.067 km/h