Question

Robbers in a car travelling at 20 m/s pass a policeman on a motorcycle at rest. The policeman immediately starts chasing the robbers. The policeman accelerates at 3 m/s square for 12 seconds and thereafter travels at a constant velocity. Calculate the distance covered by the policeman before he overtakes the car​

Answer 1

Answer :-

$\:\\\large{\underbrace{\underline{\sf{What\;the\;Question\;Says\;:-}}}}$

Here the concept of Equations of Motion has been used. Firstly, we can find out distance travelled by Policeman in 12 seconds. Then we can find out his velocity. After finding this we need to calculate distance travelled by him in 12 + t seconds so that he overtakes the robber. Also distance of robber will also be same by that.

Here,

• initial velocity = u

• final velocity = v

• acceleration = a

• time taken = t

• displacement = s

Let's do it  !!

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★ Formula Used :-

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

$\:\\\large{\boxed{\sf{\odot\;\;v\;+\;u\;=\;\bf{at}}}}$

$\:\\\large{\boxed{\sf{\odot\;\;Distance\;=\;\bf{Speed\:\times\:Time}}}}$

$\:\large{\sf{\odot\;\;Distance\;of\;robber_{(in\;(12\;+\;t))\;seconds}\;=\;\bf{Distance\;of\;Policeman_{(in\;(12\;+\;t))\;seconds}}}}$

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★ Solution :-

Given,

» Velocity of robbers in the car = 20 m/sec

» Acceleration of policeman = 3 m/sec²

» Time taken by policeman = 12 seconds

» Initial velocity of Policeman = 0 m/sec

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~ For the distance covered by Policeman in 12 seconds :-

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

$\:\\\large{\sf{:\longrightarrow\;\;\;s\;=\;\bf{(0\:\times\:12)\;+\;\dfrac{1}{2}\:\times\:3\:\times\:(12)^{2}}}}$

$\:\\\large{\sf{:\longrightarrow\;\;\;s\;=\;\bf{0\;+\;\dfrac{1}{2}\:\times\:3\:\times\:144\:\;=\:\;\underline{\underline{216\;\;m}}}}}$

$\;\\\large{\boxed{\boxed{\tt{Distance\;\;covered\;\;by\;\;Policeman\;\;in\;\;12\;\;seconds\;=\;\bf{216\;\;m}}}}}$

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~ For velocity of the policeman during this distance :-

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\;v\;+\;u\;=\;\bf{at}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\;v_{(Policeman)}\;+\;0\;=\;\bf{3\:\times\:12}}}$

$\:\\\qquad\large{\sf{:\longrightarrow\;\;\;v_{(Policeman)}\;\;=\;\underline{\underline{\bf{36\;\;m\:sec^{-1}}}}}}$

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~ For the distance of Policeman and Robber in t seconds :-

Since, policeman has constant velocity after 12 seconds, so velocity will be 36t only.

$\:\\\large{\sf{\rightarrow\;\;\;Distance\;=\;\bf{Speed\:\times\:Time}}}$

We are given that policeman already covers 216 m in 12 seconds. Now we need to find his distance in t seconds. So,

$\:\\\sf{\rightarrow\;\;\;Distance\;=\;\bf{36\:\times\:t\;\:=\;\:36t}}$

$\:\\\large{\sf{\rightarrow\;\;\;Distance\;covered \;by\;Policeman\;in\;(12\;+\;t)\;seconds\;=\;\underline{\underline{\bf{216\;+\;36t}}}}}$

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Also we are given the velocity of the robber that is 20 m/sec. Now distance coveres by him in (12 + t) seconds is :-

$\:\\\sf{\rightarrow\;\;\;Distance\;=\;\bf{20\:\times\:(12\;+\;t)\;\:=\;\:240\;+\;20t}}$

$\:\\\large{\sf{\rightarrow\;\;\;Distance\;covered \;by\;Robber\;in\;(12\;+\;t)\;seconds\;=\;\underline{\underline{\bf{240\;+\;20t}}}}}$

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~ For the Value of t :-

$\:\large{\sf{:\Longrightarrow\;\;\;Distance\;of\;robber_{(in\;(12\;+\;t))\;seconds}\;=\;\bf{Distance\;of\;Policeman_{(in\;(12\;+\;t))\;seconds}}}}$

$\:\\\large{\sf{:\Longrightarrow\;\;\;240\;+\;20t\;=\;\bf{216\;+\;36t}}}$

$\:\\\large{\sf{:\Longrightarrow\;\;\;36t\;-\;20t\;=\;\bf{240\;-\;216}}}$

$\:\\\large{\sf{:\Longrightarrow\;\;\;16t\;=\;\bf{24}}}$

$\:\\\large{\sf{:\Longrightarrow\;\;\;t\;=\;\bf{\dfrac{24}{16}\;\:=\;\;\underline{\underline{1.5\;\;seconds}}}}}$

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~ For Distance covered by Policeman before he overtakes the car :-

$\:\large{\sf{:\longmapsto\;\;\;Distance\;of\;Policeman\;to\;overtake\;car\;=\;\bf{Distance\;of\;Policeman\;in\;(12\;+\;t))\;seconds}}}$

$\:\\\large{\sf{:\longmapsto\;\;\;Distance\;covered\;by\;Policeman\;to\;overtake\;car\;=\;\bf{216\;+\;36t}}}$

$\:\\\large{\sf{:\longmapsto\;\;\;Distance\;covered\;by\;Policeman\;to\;overtake\;car\;=\;\bf{216\;+\;36(1.5)}}}$

$\:\\\large{\sf{:\longmapsto\;\;\;Distance\;covered\;by\;Policeman\;to\;overtake\;car\;=\;\bf{216\;+\;54\;\:=\;\:\underline{\underline{270\;\;m}}}}}$

$\:\\\large{\underline{\underline{\rm{Thus\;distance\;covered\;by\;policeman\;before\;he\;overtakes\;the\;car\;is\;\;\boxed{\bf{270\;\;m}}}}}}$

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★ More to know :-

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

$\:\\\sf{\leadsto\;\;\;acceleration\;=\;\dfrac{v\;-\;u}{t}}$

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

$\:\\\sf{\leadsto\;\;\;s\;=\;\dfrac{u\;+\;v}{2}\:\times\:t}$

$\:\\\sf{\leadsto\;\;\;s\;=\;vt\;-\;\dfrac{1}{2}at^{2}}$

Answer 2

Given,

Robbers in a car travelling at 20 m/s pass a policeman on a motorcycle at rest. The policeman immediately starts chasing the robbers. The policeman accelerates at 3 m/s square for 12 seconds and thereafter travels at a constant velocity.

To find :

Calculate the distance covered by the policeman before he overtakes the car​.

Solution :

In the case of policeman,

Initial velocity, u = 0 m/s [As it was at rest]

Acceleration, a = 3 m/s²

Time, t = 12 s

Now using 2nd equation of motion :

⇒ s = ut + 1/2 at²

⇒ s = 0 * 12 + 1/2 * 3 * 12²

⇒ s = 0 + 1/2 * 3 * 144

⇒ s = 432/2

⇒ s = 216 m

∴ Distance covered by policeman while constant acceleration = 216 m

Now while constant acceleration , final velocity :

⇒ v = u + at

⇒ v = 0 + 3 * 12

⇒ v = 36 m/s

∴ Distance covered = Speed * Time  [While constant velocity]

⇒ Distance covered = 36 * t

⇒ Distance covered = 36t

∴ Total distance covered by policeman = (216 + 36t) m

Now, policeman will overtake the robbers, so :

Distance covered by robbers = Distance covered by policeman.

Now time taken by the robbers = (12 + t) s

Speed of robbers = 20 m/s

∴ Distance covered by robbers = 20(12 + t) m

So atq,

⇒ 216 + 36t = 20(12 + t)

⇒ 216 + 36t = 240 + 20t

⇒ 36t - 20t = 240 - 216

⇒ 16t = 24

⇒ t = 24/16

⇒ t = 1.5 s

Now total distance covered by policeman :

⇒ Total distance covered by policeman = 216 + 36 * 1.5

⇒ Total distance covered by policeman = 216 + 54

⇒ Total distance covered by policeman = 270 m

∴ Distance covered by policeman before he overtakes the robbers' car = 270 m.