
The position coordinate of a moving particle is given by $x=6+18t+9t^{2}$ (x in meters and t in seconds). What is its velocity at $t=2sec$?
Answer
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Hint: We know that the rate of change of displacement is called velocity and the rate of change of velocity is called acceleration. Here, instead a value for displacement and time, an equation is given. So, we need to differentiate the equation to find the velocity .
Formula used:
$v=\dfrac{displacement}{time}$ Or $v=\dfrac{dx}{dt}$
Complete step by step answer:
We know that the velocity $v$ i.e. $v=\dfrac{displacement}{time}$. Or $v=\dfrac{dx}{dt}$ where time is $t$ and displacement is $x$.
Here, it is given that$x=6+18t+9t^{2}$, clearly, $x$ is given in term of $t$ .Here, using the mathematical
differentiation of $x^{n}$, then $\dfrac{d}{dx}x^{n}=nx^{n-1}$, then velocity$v=\dfrac{dx}{dt}=18+18t$
Since we also know that $\dfrac{d}{dx}K=0$ where $K$ is a constant and is independent of $x$, thus the $6$, here in the equation vanishes in the velocity equation.
Then at $t=2sec$, velocity is given as $v=18+18\times 2=18\times 3=54m/s$
Hence the velocity at $t=2sec$ is $54m/s$
Additional information:
Similarly, the acceleration$a$ is defined as the rate of change of velocity i.e. $a=\dfrac{velocity}{time}$. Or $a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$ where, time is $t$ and velocity$v$. Also see that$\dfrac{dx}{dt}=\dfrac{1}{\dfrac{dt}{dx}}$, this is the most important step in this question. Also note that$v=\dfrac{displacement}{time}$ Or $v=\dfrac{dx}{dt}$ and $a=\dfrac{velocity}{time}$.Or $a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$. To calculate, $a$ we must differentiate only $v$ with respect to $t$ and not $\dfrac{dt}{dx}$. However, acceleration is not asked in the question, but it is good to know. One must be aware of basic differentiation to solve such sums.
Note:
This may seem as a hard question at first. But this question is easy, provided you know differentiation, here we use the mathematical differentiation of $x^{n}$, then $\dfrac{d}{dx}x^{n}=nx^{n-1}$, here, in our sum, $n=-1$. And using chain rule of differentiation, we get the result.
Formula used:
$v=\dfrac{displacement}{time}$ Or $v=\dfrac{dx}{dt}$
Complete step by step answer:
We know that the velocity $v$ i.e. $v=\dfrac{displacement}{time}$. Or $v=\dfrac{dx}{dt}$ where time is $t$ and displacement is $x$.
Here, it is given that$x=6+18t+9t^{2}$, clearly, $x$ is given in term of $t$ .Here, using the mathematical
differentiation of $x^{n}$, then $\dfrac{d}{dx}x^{n}=nx^{n-1}$, then velocity$v=\dfrac{dx}{dt}=18+18t$
Since we also know that $\dfrac{d}{dx}K=0$ where $K$ is a constant and is independent of $x$, thus the $6$, here in the equation vanishes in the velocity equation.
Then at $t=2sec$, velocity is given as $v=18+18\times 2=18\times 3=54m/s$
Hence the velocity at $t=2sec$ is $54m/s$
Additional information:
Similarly, the acceleration$a$ is defined as the rate of change of velocity i.e. $a=\dfrac{velocity}{time}$. Or $a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$ where, time is $t$ and velocity$v$. Also see that$\dfrac{dx}{dt}=\dfrac{1}{\dfrac{dt}{dx}}$, this is the most important step in this question. Also note that$v=\dfrac{displacement}{time}$ Or $v=\dfrac{dx}{dt}$ and $a=\dfrac{velocity}{time}$.Or $a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$. To calculate, $a$ we must differentiate only $v$ with respect to $t$ and not $\dfrac{dt}{dx}$. However, acceleration is not asked in the question, but it is good to know. One must be aware of basic differentiation to solve such sums.
Note:
This may seem as a hard question at first. But this question is easy, provided you know differentiation, here we use the mathematical differentiation of $x^{n}$, then $\dfrac{d}{dx}x^{n}=nx^{n-1}$, here, in our sum, $n=-1$. And using chain rule of differentiation, we get the result.
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