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Question: The position coordinate of a moving particle is given by \(x=6+18t+9t^{2}\) (x in meters and t in se...

The position coordinate of a moving particle is given by x=6+18t+9t2x=6+18t+9t^{2} (x in meters and t in seconds). What is its velocity at t=2sect=2sec?

Explanation

Solution

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=displacementtimev=\dfrac{displacement}{time} Or v=dxdtv=\dfrac{dx}{dt}

Complete step by step answer:
We know that the velocity vv i.e. v=displacementtimev=\dfrac{displacement}{time}. Or v=dxdtv=\dfrac{dx}{dt} where time is tt and displacement is xx.
Here, it is given thatx=6+18t+9t2x=6+18t+9t^{2}, clearly, xx is given in term of tt .Here, using the mathematical
differentiation of xnx^{n}, then ddxxn=nxn1\dfrac{d}{dx}x^{n}=nx^{n-1}, then velocityv=dxdt=18+18tv=\dfrac{dx}{dt}=18+18t
Since we also know that ddxK=0\dfrac{d}{dx}K=0 where KK is a constant and is independent of xx, thus the 66, here in the equation vanishes in the velocity equation.
Then at t=2sect=2sec, velocity is given as v=18+18×2=18×3=54m/sv=18+18\times 2=18\times 3=54m/s
Hence the velocity at t=2sect=2sec is 54m/s54m/s

Additional information:
Similarly, the accelerationaa is defined as the rate of change of velocity i.e. a=velocitytimea=\dfrac{velocity}{time}. Or a=dvdt=d2xdt2a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}} where, time is tt and velocityvv. Also see thatdxdt=1dtdx\dfrac{dx}{dt}=\dfrac{1}{\dfrac{dt}{dx}}, this is the most important step in this question. Also note thatv=displacementtimev=\dfrac{displacement}{time} Or v=dxdtv=\dfrac{dx}{dt} and a=velocitytimea=\dfrac{velocity}{time}.Or a=dvdt=d2xdt2a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}. To calculate, aa we must differentiate only vv with respect to tt and not dtdx\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 xnx^{n}, then ddxxn=nxn1\dfrac{d}{dx}x^{n}=nx^{n-1}, here, in our sum, n=1n=-1. And using chain rule of differentiation, we get the result.