Question

If a car covers 2/5th of the total distance with v1 speed and 3/5th distance with v2 then average speed is...​

Answer 1

Answer:

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Answer 2

★ Concept

Here the concept of Distance, Time and Speed relationship has been used. We see that the body covers different part of total distance with different speed. So firstly we can find the time taken to cover that part of total distance with that Speed. Then by adding both we can get total time to cover the total distance. And then we shall apply the formula of average speed and find the answer.

Let's do it !!

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★ Formula Used :-

$\;\boxed{\sf{\pink{Time\;=\;\bf{\dfrac{Distance}{Velocity}}}}}$

$\;\boxed{\sf{\pink{V_{av}\;=\;\bf{\dfrac{Total\;Distance}{Total\;Time}}}}}$

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★ Solution :-

Given,

» Speed to cover ⅖ of total distance = v₁

» Speed to cover ⅗ of total distance = v₂

• Let the total distance be s

• Let the time taken to cover first part of journey that is ⅖ s = t₁

• Let the time taken to cover second part of journey that is ⅗ s = t₂

• Total time taken to cover journey = T = t₁ + t₂

Then,

• Speed to cover ⅖ s = v₁

• Speed to cover ⅗ s = v₂

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~ For the value of t₁ and t₂ ::

We know that,

$\;\sf{\rightarrow\;\;Time\;=\;\bf{\dfrac{Distance}{Velocity}}}$

• For t₁ ::

By applying the values in the formula, we get

$\;\sf{\rightarrow\;\;t_{1}\;=\;\bf{\dfrac{\dfrac{2}{5}\:\times\:s}{v_{1}}}}$

$\;\sf{\rightarrow\;\;\green{t_{1}\;=\;\bf{\dfrac{2s}{5v_{1}}}}}$

• For t₂ ::

By applying the values in the formula, we get

$\;\sf{\rightarrow\;\;t_{2}\;=\;\bf{\dfrac{\dfrac{3}{5}\:\times\:s}{v_{2}}}}$

$\;\sf{\rightarrow\;\;\orange{t_{2}\;=\;\bf{\dfrac{3s}{5v_{2}}}}}$

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~ For the average speed of the car ::

• Let average speed be Vav

We know that,

$\;\sf{\Longrightarrow\;\;V_{av}\;=\;\bf{\dfrac{Total\;Distance}{Total\;Time}}}$

By applying values, we get

$\;\sf{\Longrightarrow\;\;V_{av}\;=\;\bf{\dfrac{s}{T}}}$

$\;\sf{\Longrightarrow\;\;V_{av}\;=\;\bf{\dfrac{s}{t_{1}\:+\:t_{2}}}}$

$\;\sf{\Longrightarrow\;\;V_{av}\;=\;\bf{\dfrac{s}{\dfrac{2s}{5v_{1}}\:+\:\dfrac{3s}{5v_{2}}}}}$

$\;\sf{\Longrightarrow\;\;V_{av}\;=\;\bf{\dfrac{s}{\dfrac{2sv_{2}\:+\:3sv_{1}}{5v_{1}v_{2}}}}}$

$\;\sf{\Longrightarrow\;\;V_{av}\;=\;\bf{\dfrac{5v_{1}v_{2}\:s}{s(2v_{2}\:+\:3v_{1})}}}$

Cancelling s from both sides, we get

$\;\sf{\Longrightarrow\;\;\red{V_{av}\;=\;\bf{\dfrac{5v_{1}v_{2}}{2v_{2}\:+\:3v_{1}}}}}$

This is the required answer.

$\;\underline{\boxed{\tt{Required\;\:Answer\;=\;\bf{\purple{\dfrac{5v_{1}v_{2}}{2v_{2}\:+\:3v_{1}}}}}}}$

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★ More to know :-

$\;\tt{\leadsto\;\; Velocity\;=\;\dfrac{Displacement}{Time}}$

$\;\tt{\leadsto\;\;Acceleration\;=\;\dfrac{\Delta V}{t}}$

$\;\tt{\leadsto\;\;v\;-\;u\;=\;at}$

$\;\tt{\leadsto\;\;s\;=\;ut\;+\;\dfrac{1}{2}at^{2}}$

$\;\tt{\leadsto\;\;v^{2}\;-\;u^{2}\;=\;2as}$