Question

Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.​

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Answer 1

Answer:

Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.

Using the equation of motion s = ut + at 1/ 2 at2

Distance travelled in 5 s

s = u × 5 +1/2 a × 52

or s = 5 u + 25/ 2 a ——(i)

Similarly, distance travelled in 4 s, s′ = 4 u + 16/ 2 a——(ii)

Distance travelled in the interval between 4th and 5th second

= (s – s′) = (u +9/2 a)m

Answer 2

Answer:

$\\\ \: \underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Second Equation of Motion has been used . We see that we have to find the relation of distance covered between 4th and 5th second with same acceleration and same initial velocity . So Position - Time Relation will be best for it .

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{s\;=\;\bf{ut\;+\;\dfrac{1}{2}\:at^{2}}}}}$

This is the Second Equation of Motion commonly known as Position - Time Relationship .

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★ Solution :-

Given,

» Initial velocity of the body = u

» Acceleration of the body = a

» Time taken by the body for first case = T₁ = 4 second

» Tine taken by the body for second case = T₂ = 5 second

» Distance covered in 4th second = s₄

» Distance covered in 5th second = s₅

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~ Distance covered in 4th second ::

We know that,

$\\\;\sf{:\rightarrow\;\;s\;=\;\bf{ut\;+\;\dfrac{1}{2}\:at^{2}}}$

By applying values, we get,

$\\\;\sf{:\Longrightarrow\;\;s_{4}\;=\;\bf{uT_{1}\;+\;\dfrac{1}{2}\:aT_{1} ^{2}}}$

$\\\;\sf{:\Longrightarrow\;\;s_{4}\;=\;\bf{u(4)\;+\;\dfrac{1}{2}\:a(4)^{2}}}$

$\\\;\bf{:\Longrightarrow\;\;s_{4}\;=\;\bf{\green{4u\;+\;\dfrac{16}{2}\:a}}}$

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~ Distance covered in 5th second ::

We know that,

$\\\;\sf{:\rightarrow\;\;s\;=\;\bf{ut\;+\;\dfrac{1}{2}\:at^{2}}}$

By applying values, we get,

$\\\;\sf{:\Longrightarrow\;\;s_{5}\;=\;\bf{uT_{2}\;+\;\dfrac{1}{2}\:aT_{2} ^{2}}}$

$\\\;\sf{:\Longrightarrow\;\;s_{5}\;=\;\bf{u(5)\;+\;\dfrac{1}{2}\:a(5)^{2}}}$

$\\\;\bf{:\Longrightarrow\;\;s_{4}\;=\;\bf{\orange{5u\;+\;\dfrac{25}{2}\:a}}}$

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~ Distance covered between 4th and 5th second ::

This is given as difference between the distance covered in 5th second and distance covered in 4th second.

$\\\;\sf{\mapsto\;\;Required\;Distance\;=\;\bf{\blue{s_{5}\;-\;s_{4}}}}$

Then,

$\\\;\sf{\mapsto\;\;s_{5}\;-\;s_{4}\;=\;\bf{\bigg(5u\;+\;\dfrac{25}{2}\:a\bigg)\;-\;\bigg(4u\;+\;\dfrac{16}{2}\:a\bigg)}}$

$\\\;\sf{\mapsto\;\;s_{5}\;-\;s_{4}\;=\;\bf{5u\;+\;\dfrac{25\:a}{2}\;-\;4u\;-\;\dfrac{16\:a}{2}}}$

$\\\;\sf{\mapsto\;\;s_{5}\;-\;s_{4}\;=\;\bf{5u\;-\;4u\;+\;\bigg(\dfrac{25\:a}{2}\;-\;\dfrac{16\:a}{2}\bigg)}}$

$\\\;\sf{\mapsto\;\;s_{5}\;-\;s_{4}\;=\;\bf{u\;+\;\bigg(\dfrac{25\:a\;-\;16\:a}{2}\bigg)}}$

$\\\;\sf{\mapsto\;\;s_{5}\;-\;s_{4}\;=\;\bf{u\;+\;\bigg(\dfrac{9\:a}{2}\bigg)}}$

$\\\;\sf{\mapsto\;\;s_{5}\;-\;s_{4}\;=\;\bf{\red{u\;+\;\dfrac{9\:a}{2}}}}$

Thus the relation which defines the distance travelled between 5th and 4th second.

$\\\;\underline{\boxed{\tt{Required\;\:Relation\;=\;\bf{\purple{u\;+\;\dfrac{9\:a}{2}}}}}}$

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★ More to know :-

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

This is the First Equation of Motion which is also known as Velocity - Time Relation.

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

This is the Third Equation of Motion which is also known as Velocity - Position Relation.

$\\\;\sf{\leadsto\;\;s_{n_{th}}\;=\;u\;+\;\dfrac{a}{2}\:(2n\;-\;1)}$

This is the Fourth Equation of Motion which is used to find the value at particular time.