
The velocity of a particle moving in a positive direction of the x-axis varies as $v = \alpha \sqrt x $, where a is a positive constant. If at $t = 0$ , $x = 0$, the velocity and acceleration as a function of time $t$ will be.
A. \[\dfrac{{{\alpha ^2}t}}{2},\,{\alpha ^2}\]
B. \[{\alpha ^2}t,\,\dfrac{{{\alpha ^2}t}}{2}\]
C. \[\dfrac{{{\alpha ^2}t}}{2},\,\dfrac{{{\alpha ^2}}}{2}\]
D. \[{\alpha ^2}t,\,{\alpha ^2}\]
Answer
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Hint: The difference between the final and initial velocity with respect to time is called acceleration. And it is denoted by ‘a’ and SI unit is $m{s^{ - 2}}$. Acceleration is a vector quantity, it means it has direction as well as magnitude. There are two types of acceleration: uniform acceleration and non-uniform accelerations.
Complete step by step answer:
Given, $v = \alpha \sqrt x $…(1)
We know that
$\dfrac{{dV}}{{dT}} = a$…(2)
Differentiate equation 1 w.r.t to T
$\dfrac{{dV}}{{dT}} = \alpha \dfrac{d}{{dT}}\sqrt x $
$ \Rightarrow \dfrac{{dV}}{{dT}} = \dfrac{\alpha }{{2\sqrt x }}\dfrac{{dx}}{{dT}}$
But $\dfrac{{dx}}{{dT}} = v$ is velocity gradient
Put the values in equation 2 we get.
$ \Rightarrow a = \dfrac{\alpha }{{2\sqrt x }} \times \alpha \sqrt x $
$ \Rightarrow a = \dfrac{{{\alpha ^2}}}{2}$
Now, relationship between velocity and displacement
$v = \alpha \sqrt x $
$ \Rightarrow \dfrac{{dx}}{{dt}} = \alpha \sqrt x $
$ \Rightarrow \dfrac{{dx}}{{\sqrt x }} = \alpha dt$
Integrate
$ \Rightarrow \int {\dfrac{{dx}}{{\sqrt x }}} = \alpha \int {dt} $
$ \Rightarrow \dfrac{{{x^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}} = \alpha t + c$
Since $t = 0,\,x = 0$
We get
$c = 0$
So, $2\sqrt x = \alpha t$
Squaring both side
$ \Rightarrow 4x = {a^2}{t^2}$
$ \Rightarrow x = \dfrac{{{\alpha ^2}{t^2}}}{4}$
Differentiate w.r.t. ‘t’
$ \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{{{\alpha ^2}}}{4}\dfrac{d}{{dt}}({t^2})$
We know that
$ \dfrac{{dx}}{{dt}} = v$
$ \therefore v = \dfrac{{{\alpha ^2}t}}{2}$
Hence, the correct answer is option C.
Note: Force is the product of mass and acceleration. Application acceleration improves the application performances using techniques like compression, caching and transmission control protocol. When the velocity of any moving is changed it is said to be a change in acceleration. Change in acceleration may be positive or negative.
Complete step by step answer:
Given, $v = \alpha \sqrt x $…(1)
We know that
$\dfrac{{dV}}{{dT}} = a$…(2)
Differentiate equation 1 w.r.t to T
$\dfrac{{dV}}{{dT}} = \alpha \dfrac{d}{{dT}}\sqrt x $
$ \Rightarrow \dfrac{{dV}}{{dT}} = \dfrac{\alpha }{{2\sqrt x }}\dfrac{{dx}}{{dT}}$
But $\dfrac{{dx}}{{dT}} = v$ is velocity gradient
Put the values in equation 2 we get.
$ \Rightarrow a = \dfrac{\alpha }{{2\sqrt x }} \times \alpha \sqrt x $
$ \Rightarrow a = \dfrac{{{\alpha ^2}}}{2}$
Now, relationship between velocity and displacement
$v = \alpha \sqrt x $
$ \Rightarrow \dfrac{{dx}}{{dt}} = \alpha \sqrt x $
$ \Rightarrow \dfrac{{dx}}{{\sqrt x }} = \alpha dt$
Integrate
$ \Rightarrow \int {\dfrac{{dx}}{{\sqrt x }}} = \alpha \int {dt} $
$ \Rightarrow \dfrac{{{x^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}} = \alpha t + c$
Since $t = 0,\,x = 0$
We get
$c = 0$
So, $2\sqrt x = \alpha t$
Squaring both side
$ \Rightarrow 4x = {a^2}{t^2}$
$ \Rightarrow x = \dfrac{{{\alpha ^2}{t^2}}}{4}$
Differentiate w.r.t. ‘t’
$ \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{{{\alpha ^2}}}{4}\dfrac{d}{{dt}}({t^2})$
We know that
$ \dfrac{{dx}}{{dt}} = v$
$ \therefore v = \dfrac{{{\alpha ^2}t}}{2}$
Hence, the correct answer is option C.
Note: Force is the product of mass and acceleration. Application acceleration improves the application performances using techniques like compression, caching and transmission control protocol. When the velocity of any moving is changed it is said to be a change in acceleration. Change in acceleration may be positive or negative.
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