Question

Hello Brainlians !!
Here is a question to check your Kinematics.
Please don't spam. No spams are allowed.

An electron travelling with a speed of 5 × 10³ m/sec passes through an electric field with acceleration of 10¹² m/sec.
i) How long will it take for the electron to double its speed ?

ii) What will be the distance covered by the electron in this time ?

Don't be greedy for points. Explain in detail. No copied answers are needed. Only good users answer it.​

Answer 1

Answer :-

$\:\\\large{\underbrace{\underline{\sf{Let's\;understand\;the\;Question\;:-}}}}$

Here the concept of Acceleration - Time and Second Equation of Motion has been used. We know that acceleration is the rate of change of velocity. Thus by applying values in this, we can find our first answer. And then in Second Equation of Motion we can apply values and find our second answer.

Let's do it !!

_______________________________________________

★ Formula Used :-

$\:\\\large{\boxed{\sf{a\;=\;\bf{\dfrac{v\;-\;u}{t}}}}}$

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

_______________________________________________

★ Solution :-

Given,

» Initial Velocity of electron = u = 5 × 10³ m/sec

» Acceleration done by electron = a = 10¹² m/sec

Since, the electron doubles its speed after sometime. Thus, electron has

» Final Velocity = v = 2 × 5 × 10³ m/sec

_______________________________________________

i.) ~ For Time Taken by the Electron :-

• Let the time taken be denoted by 't'. Then,

By using the relation of acceleration and time, we get,

$\\\:\;\;\;\;\large{\sf{:\Longrightarrow\;\;\; a\;=\;\bf{\dfrac{v\;-\;u}{t}}}}$

$\\\:\;\;\;\;\large{\sf{:\Longrightarrow\;\;\;t\;=\;\bf{\dfrac{v\;-\;u}{a}}}}$

$\\\:\;\;\;\;\large{\sf{:\Longrightarrow\;\;\;t\;=\;\bf{\dfrac{(2\:\times\:5\:\times\:10^{3})\;-\;(5\:\times\:10^{3})}{10^{12}}}}}$

$\\\:\;\;\;\;\large{\sf{:\Longrightarrow\;\;\;t\;=\;\bf{\dfrac{(10\:\times\:10^{3})\;-\;(5\:\times\:10^{3})}{10^{12}}}}}$

$\\\:\;\;\;\;\large{\sf{:\Longrightarrow\;\;\;t\;=\;\bf{\dfrac{(5\:\times\:10^{3})}{10^{12}}}}}$

$\\\:\;\;\;\;\large{\sf{:\Longrightarrow\;\;\;t\;=\;\underline{\underline{\bf{(5\:\times\:10^{-\:9})\;s}}}}}$

$\\\;\large{\underline{\underline{\rm{Thus,\;time\;taken\;for\;the\;electron\;to\;double\;its\;speed\;is\;\;\boxed{\bf{5\:\times\:10^{-\:9}\;s}}}}}}$

_______________________________________________

ii.) ~ For the Distance coveres by electron in this time :-

Let the distance covered by electron be 's'.

Then,

We already have the required values. Now we can apply them in the Second Equation of Motion. Then,

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

$\\\:\;\;\;\;\large{\sf{:\mapsto\;\;\;s\;=\;\bf{(5\;\times\;10^{3}\;\times\;5\;\times\;10^{-\:9})\;+\;\dfrac{1}{2}\;\times\;10^{12}\;\times\;(5\;\times\;10^{-\:9})^{2}}}}$

$\\\:\;\;\;\;\large{\sf{:\mapsto\;\;\;s\;=\;\bf{(5\;\times\;10^{3}\;\times\;5\;\times\;10^{-\:9})\;+\;\dfrac{1}{2}\;\times\;10^{12}\;\times\;(25\;\times\;10^{-\:18})}}}$

$\\\:\;\;\;\;\large{\sf{:\mapsto\;\;\;s\;=\;\bf{(5\;\times\;10^{3}\;\times\;5\;\times\;10^{-\:9})\;+\;10^{12}\;\times\;(12.5\;\times\;10^{-\:18})}}}$

$\\\:\;\;\;\;\large{\sf{:\mapsto\;\;\;s\;=\;\bf{25\;\times\;10^{-\:6}\;+\;(12.5\;\times\;10^{-\:6})}}}$

$\\\:\;\;\;\;\large{\sf{:\mapsto\;\;\;s\;=\;\bf{(25\;+\;12.5)\;\times\;10^{-\:6}}}}$

$\\\:\;\;\;\;\large{\sf{:\mapsto\;\;\;s\;=\;\underline{\underline{\bf{37.5\;\times\;10^{-\:6}\;m}}}}}$

$\\\;\large{\underline{\underline{\rm{Thus,\;distance\;covered\;by\;electron\;in\;this\;time\;is\;\;\boxed{\bf{37.5\;\times\;10^{-\:6}\;m}}}}}}$

_______________________________________________

★ Let's know some more Formulas :-

$\\\:\sf{\leadsto\;\;\;v \;+\;u\;=\;at}$

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

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

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

$\\\:\sf{\leadsto\;\;\;Maximum\;Height,\;h\;=\;\dfrac{u^{2}}{g}}$

$\\\:\sf{\leadsto\;\;\;Time_{(ascent\:or\:descent)},\;t\;=\;\dfrac{u}{g}}$

$\\\:\sf{\leadsto\;\;\;Time\;of\;Flight,\;t\;=\;\dfrac{2u}{g}}$

$\\\:\sf{\leadsto\;\;\;v_{(u\:=\:0)}\;=\;\sqrt{2gh}}$

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

Answer 2

Answer:

this is your answer hope it helps you