Question

Find time when boy catches the girl or when they are at their closest separation..
velocity of boy is 50m/s and velocity of girl is 30m/s and acceleration of boy is 1m/s² and acceleration of girl is 2m/s² and distance between them is 150m​.

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Answer 1

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

Here the Concept of Relative Velocity and Relative Acceleration as well as Third Equation of Motion has been used. We see that we are given the velocities and acceleration of both girl and boy. Even its given the distance between the boy and the girl. And we need to find the time taken by boy to reach the nearest position of catch the girl. Here we shall take the nearest position as where the boy catches girl. Firstly we shall find the relative velocities of boy and girl. Then we shall find relative acceleration and then using these, we can find the time taken.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{Relative\;Velocity,\;v_{1\:2}\;=\;\bf{v_{1}\;-\;v_{2}}}}}$

This shows relative velocity of 1 with respect to 2.

$\\\;\boxed{\sf{\pink{Relative\;Acceleration,\;a_{1\:2}\;=\;\bf{a_{1}\;-\;a_{2}}}}}$

This shows relative velocity of 1 with respect to 2.

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

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

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★ Solution :-

Given,

» Velocity of Boy = v₁ = 50 m/sec

» Velocity of Girl = v₂ = 30 m/sec

» Acceleration of Boy = a₁ = 1 m/sec²

» Acceleration of Girl = a₂ = 2 m/sec

We are already given to find that when boy catches the girl. So here all the relative motion will be Boy with respect to Girl.

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~ For relative velocity of Boy with respect to Girl ::

• Let the relative velocity of Boy with respect to girl be v₁₂

By using the formula of relative velocity,

$\\\;\sf{:\rightarrow\;\;Relative\;Velocity,\;v_{1\:2}\;=\;\bf{v_{1}\;-\;v_{2}}}$

By applying values, we get

$\\\;\sf{:\rightarrow\;\;Relative\;Velocity,\;v_{1\:2}\;=\;\bf{50\;-\;30}}$

$\\\;\bf{:\rightarrow\;\;Relative\;Velocity,\;v_{1\:2}\;=\;\bf{\orange{20\;\;m\:sec^{-1}}}}$

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~ For relative acceleration of Boy with respect to Girl ::

• Let the relative acceleration of Boy with respect to Girl be a₁₂

Then by using the formula of relative acceleration, we get

$\\\;\sf{:\rightarrow\;\;Relative\;Acceleration,\;a_{1\:2}\;=\;\bf{a_{1}\;-\;a_{2}}}$

By applying values, we get

$\\\;\sf{:\rightarrow\;\;Relative\;Acceleration,\;a_{1\:2}\;=\;\bf{1\;-\;2}}$

$\\\;\bf{:\rightarrow\;\;Relative\;Acceleration,\;a_{1\:2}\;=\;\bf{\blue{-\:1\;\:m\:sec^{-2}}}}$

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~ For Time Taken by Boy to catch the Girl ::

According to the second equation of motion, we know that

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

• Let the required time be t seconds.

Now by applying values, we get

$\\\;\sf{:\Longrightarrow\;\;150\;=\;\bf{(20)t\;+\;\dfrac{1}{2}\:(-1)t^{2}}}$

• Since here u denotes velocity, so we are using relative velocity.

$\\\;\sf{:\Longrightarrow\;\;150\;=\;\bf{20\:t\;-\;\dfrac{t^{2}}{2}}}$

Now taking 2 as common denominator, we get

$\\\;\sf{:\Longrightarrow\;\;150\;=\;\bf{\dfrac{40\:t}{2}\;-\;\dfrac{t^{2}}{2}}}$

Transposing 2 to other side, we get

$\\\;\sf{:\Longrightarrow\;\;2\:\times\:150\;=\;\bf{40\:t\;-\;t^{2}}}$

$\\\;\sf{:\Longrightarrow\;\;300\;=\;\bf{40\:t\;-\;t^{2}}}$

$\\\;\bf{:\Longrightarrow\;\;\red{t^{2}\;-\;40t\;+\;300\;=\;0}}$

This forms a Quadratic Equation. Easiest way to solve this is Splitting the Middle Term. Then, we get

$\\\;\bf{:\Longrightarrow\;\;t^{2}\;-\;40t\;+\;300\;=\;0}$

This can be written as, since (30 × 10 = 300)

$\\\;\sf{\gray{:\Longrightarrow\;\;t^{2}\;-\;30t\;-\;10t\;+\;300\;=\;0}}$

• Taking the common terms, we get

$\\\;\sf{\gray{:\Longrightarrow\;\;t\bf{(t\;-\;30)}\;-\;10\bf{(t\;-\;30)}\;=\;0}}$

Now combining the common terms, we get

$\\\;\sf{\gray{:\Longrightarrow\;\;(t\;-\;30)(t\;-\;10)\;=\;0}}$

Here either (t - 30) = 0 or (t - 10) = 0

Then,

✒ (t - 30) = 0 and (t - 10) = 0

✒ t = 30 and t = 10

Clearly here we get two values of time. This means there will be two point of time when boy will be able to catch the girl which are 30 seconds and 10 seconds.

$\\\;\underline{\boxed{\tt{Required\;\:Time\;=\;\bf{\purple{30\;seconds}}\;or\;\bf{\purple{10\;seconds}}}}}$

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★ More to know :-

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

This is First - Equation of Motion.

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

This is Third - Equation of Motion.

$\\\;\sf{\leadsto\;\;Relative\;Velocity,\;v_{2\:1}\;=\;v_{2}\;-\;v_{1}}$

$\\\;\sf{\leadsto\;\;Relative\;Acceleration,\;a_{2\:1}\;=\;a_{2}\;-\;a_{1}}$

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

This is the Fourth Equation of Motion.

Answer 2

Answer:

$\green{\tt{\therefore{Time=10\:sec\:and\:30\:sec}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\begin{gathered}\green{\underline{\bold{Given :}}} \\ \tt: \implies Velocity \: of \: boy = 50 \: m/s \\ \\ \tt: \implies Velocity \: of \: girl = 30 \: m/s \\ \\ \tt: \implies Acceleration \: of \: boy = {1 \: m/s}^{2} \\ \\ \tt: \implies Acceleration \: of \:girl= {2\: m/s}^{2} \\ \\ \tt: \implies Sepration \: between \: them = 150 \: m \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \tt: \implies Time \: taken \: to \: catch \: the \: girl =? \end{gathered}$

• According to given question :

$\begin{gathered} \green{\star} \: \text{Using \: relative \: motion \: method} \\ \\ \green{ \circ} \: \tt Net \: velocity = 50 - 30 = 20 \: m/s \\ \\ \green{ \circ} \: \tt Net \: acceleration = 1 - 2 = - 1\: m /{s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \tt: \implies 150 = 20 \times t + \frac{1}{2} \times -1 \times {t}^{2} \\ \\ \tt: \implies 300 = 40t - {t}^{2} \\ \\ \tt: \implies {t}^{2} - 40t + 300 = 0 \\ \\ \tt: \implies t = \frac{ - ( - 40) \pm\sqrt{ { (- 40)}^{2} - 4 \times 1 \times 300} }{2 \times 1} \\ \\ \tt: \implies t = \frac{40 \pm \sqrt{1600 - 1200} }{2} \\ \\ \tt: \implies t = \frac{40 \pm 20}{2} \\ \\ \green{\tt: \implies t = 10 \: sec \: and \: 30 \: sec}\end{gathered}$

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$\sf\red{@MsHeaven}$