Question
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+9t2 (x in meters and t in seconds). What is its velocity at t=2sec?
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=timedisplacement Or v=dtdx
Complete step by step answer:
We know that the velocity v i.e. v=timedisplacement. Or v=dtdx where time is t and displacement is x.
Here, it is given thatx=6+18t+9t2, clearly, x is given in term of t .Here, using the mathematical
differentiation of xn, then dxdxn=nxn−1, then velocityv=dtdx=18+18t
Since we also know that dxdK=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+18×2=18×3=54m/s
Hence the velocity at t=2sec is 54m/s
Additional information:
Similarly, the accelerationa is defined as the rate of change of velocity i.e. a=timevelocity. Or a=dtdv=dt2d2x where, time is t and velocityv. Also see thatdtdx=dxdt1, this is the most important step in this question. Also note thatv=timedisplacement Or v=dtdx and a=timevelocity.Or a=dtdv=dt2d2x. To calculate, a we must differentiate only v with respect to t and not dxdt. 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 xn, then dxdxn=nxn−1, here, in our sum, n=−1. And using chain rule of differentiation, we get the result.