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Question: The position of a particle is given by \(s = f(t) = {t^3} - 6{t^2} + 9\) where \(t\) is in seconds a...

The position of a particle is given by s=f(t)=t3−6t2+9s = f(t) = {t^3} - 6{t^2} + 9 where tt is in seconds and ss is in metres.
(a) when is the particle moving backwards (that is in the negative direction)?
(b) Find the acceleration at time t and after 4s?4s?

Explanation

Solution

In order to solve this question, we should know that the body moves backward or in negative direction only when its velocity is negative and here we will find velocity using displacement relation and then find acceleration of the particle by using derivatives of displacement and velocity function.

Complete step by step answer:
(a) According to the question, we have given the displacement and time relation as s=t3−6t2+9s = {t^3} - 6{t^2} + 9 now, if v is the velocity of the particle then, v=dsdtv = \dfrac{{ds}}{{dt}} putting displacement relation and differentiating the equation we get,
v=d(t3−6t2+9)dtv = \dfrac{{d({t^3} - 6{t^2} + 9)}}{{dt}}
Using differentiation rules ddx(xn)=nxn−1\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}} and differentiation of constant terms are zero we get,
v=3t2−12tv = 3{t^2} - 12t
Now in order to move backward or in a negative direction, velocity of the particle must be negative which shows that v<0v < 0 negative velocity. On putting value of v we get,
3t2−12t<03{t^2} - 12t < 0
∴t<4\therefore t < 4

So, the particle moves backward in the time interval of 0<t<40 < t < 4 seconds.

(b) Now, acceleration a is the derivative of velocity function so we can write acceleration aa as,
a=dvdta = \dfrac{{dv}}{{dt}}
Putting value of vv and again using differentiation rules we get,
⇒a=d(3t2−12t)dt\Rightarrow a = \dfrac{{d(3{t^2} - 12t)}}{{dt}}
⇒a=6t−12\Rightarrow a = 6t - 12
Hence, the acceleration of the particle at time t is a=6t−12a = 6t - 12 and at t=4sec⁡t = 4\sec we can calculate it as
a=6×4−12a = 6 \times 4 - 12
∴a=12 ms−2\therefore a = 12\,m{s^{ - 2}}

Hence, particle moves with an acceleration of a=12 ms−2a = 12\,m{s^{ - 2}} after 4sec⁡.4\sec .

Note: It should be remembered that, if a body is moving in a negative direction the velocity will always be negative whereas if a particle is moving forward its velocity will be positive and the derivatives of all constant functions are always zero. Remember all the differentiation rules and basic formulas while solving such questions.