Question
Answers

An electron starting from rest has a velocity that increases linearly with time. I.e $v = kt$, where $k = 2m/{s^2}$. The distance covered in the first three distances will be,
a. $36m$
b. $27m$
c. $18m$
d. $9m$

Answer
VerifiedVerified
91.5k+ views
Hint: In the question, the velocity value is given. The velocity can be determined by dividing the displacement and time. We need to use integration to solve the problem.

Formula used:
To find the velocity:
$ \Rightarrow \text{velocity} = \dfrac{\text{displacement}}{\text{time}}$
The velocity formula in terms of integration:
\[ \Rightarrow dv = \dfrac{{dx}}{{dt}}\]
Where $dv$ is the velocity, $dx$ is the displacement and $dt$ is the time.

Complete step by step answer:
We have the value of an electron that starts from the rest has the velocity of which increases linearly with the time. That is the velocity has the value of $v = kt$. We have the value of $k$. The value of $k$ is $k = 2m/{s^2}$.
To solve the given problem we can integrate the velocity to get the answer.
We have,
$ \Rightarrow v = kt$
Substitute the value of velocity. We have,
$ \Rightarrow \text{velocity} = \dfrac{\text{displacement}}{\text{time}}$

The velocity formula in terms of integration:
\[ \Rightarrow dv = \dfrac{{dx}}{{dt}}\]
Where $dv$ is the velocity, $dx$ is the displacement and $dt$ is the time.
We have,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = kt\]
Integrate on both the sides. We get,
\[ \Rightarrow \int {\dfrac{{dx}}{{dt}}} = \int {kt} \]
In left hand side we are going integrate the value of $k$ with respect to $t$. That is,
\[ \Rightarrow \int {\dfrac{{dx}}{{dt}}} = \int {kt} .dt\]

The limits in the left-hand side will be the value of $t$. The lower limit has the value of $t$ zero, as it is the initial seconds and the upper limit has the value of three.
Substitute the limits in the equation, we get,
\[ \Rightarrow \int {\dfrac{{dx}}{{dt}}} = \int\limits_{t = 0}^{t = 3} {kt} .dt\]
Integrate the equation with respect to time in both the sides we get,
\[ \Rightarrow \int {\dfrac{{dx}}{{dt}}} = \int\limits_{t = 0}^{t = 3} {kt} .dt\]
\[ \Rightarrow x = k\left[ {\dfrac{{{t^2}}}{2}} \right]_0^3\]
Substitute the limits in place of $t$ we get,
\[ \Rightarrow x = k\left[ {\dfrac{{{3^2}}}{2} - 0} \right]\]

Take square on three we get nine. That is,
\[ \Rightarrow x = k\left[ {\dfrac{9}{2} - 0} \right]\]
Substitute the value of $k$. We get,
\[ \Rightarrow x = 2\left[ {\dfrac{9}{2} - 0} \right]\]
Cancel out the common term we get,
\[ \Rightarrow x = 2 \times \dfrac{9}{2}\]
$ \Rightarrow x = 9$
Therefore, the distance covered in the first three distances will be equal to $9m$
$\therefore x = 9m$

Hence, the correct answer is option (D).

Note: While doing the integration, when we substitute the limits in the integrated part always remember that first we need to substitute the upper limit and then we need to substitute the lower limit.