Question

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Answer 1

To answer this question we need to understand the meaning of distance and displacement.

Distance = The total length covered when moving from one point to another.

Displacement = this is a vector quantity with direction and magnitude. It is the movement from one point to another in a certain direction.

If you move from one point to another and back to the same point the displacement is 0.

If she swims 200m it means she swims back to the same point and the displacement is 0.

Velocity = Displacement / time

Time = 10 × 60 = 600s

Velocity = 0/600 = 0 m/s

Distance covered = 200 m

Speed = Distance / time

Speed = 200/600 = 1/3 m/s

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Answer 2

$\underline{\underline{To\:Find}}$

$\underline{The\:average\:speed\:of\:the\:train}$

$\underline{\underline{Given:}}$

$speed\:of\:the\:train = v_{1} = 30km\:hr^{-1}$

$speed\:of\:the\:train = v_{2} = 45km\:hr^{-1}$

$\underline{\underline{We\:Know:}}$

☞ $Average\:speed = \dfrac{s_{1} + s_{2}}{t_{1} + t_{2}}$

Where;

$s = Distance\:travelled$

$t = time\:taken$

$\underline{\underline{Taken:}}$

let the distance be x in both the cases , according to the Question.

$\underline{\underline{Concept:}}$

To find the average speed , first we have to find the time taken in both the cases..

we know,

$speed = \dfrac{Distance}{Time}$

$\therefore time = \dfrac{Distance}{speed}$

Case ....(i)

$Distance = x$

$Speed = 30 km\:hr^{-1}$

let the time be $t_{1}$

$\therefore time = (\dfrac{x}{30})h$

${\boxed{\therefore t_{1} = (\dfrac{x}{30})h}}$

Case ....(ii)

$Distance = x$

$Speed = 45 km\:hr^{-1}$

let the time be $t_{2}$

$\therefore time = (\dfrac{x}{45})h$

${\boxed{\therefore t_{1} = (\dfrac{x}{45})h}}$

______________________________________

$\underline{\underline{Average\:speed}}$

We know;

$Average\:speed = \dfrac{s_{1} + s_{2}}{t_{1} + t_{2}}$

$\Rightarrow Average\:speed = \bigg(\dfrac{x + x}{\dfrac{x}{30} + \dfrac{x}{45}}\bigg)$

$\Rightarrow Average\:speed = \bigg(\dfrac{2x}{\dfrac{x}{30} + \dfrac{x}{45}}\bigg)$

LCM of denominators is 90..

$\Rightarrow Average\:speed = \bigg(\dfrac{2x}{\dfrac{2x + 3x}{90}}\bigg)$

$\Rightarrow Average\:speed = \bigg(\dfrac{2x}{\dfrac{5x}{90}}\bigg)$

$\Rightarrow Average\:speed = \dfrac{2x}{5x} × 90$

$\Rightarrow Average\:speed = \dfrac{2}{5} × 90$

$\Rightarrow Average\:speed = \dfrac{2}{1} × 18$

$\Rightarrow Average\:speed = 36km\:hr^{-1}$

${\boxed{\therefore Average\:speed = 36km\:hr^{-1}}}$

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