Question

The minimum acceleration attained on the interval $0 \leqslant t \leqslant 4$ by the particle whose velocity is given by $v(t) = {t^3} - 4{t^2} - 3t + 2$ is?

Answer 1

To solve this type of question, one must know the basics of differentiation. We will simply apply differentiation as we know that the rate of change of velocity is acceleration and we will simply put the obtained value of time in the equation to obtain the final answer.

Complete step by step answer:
According to the question it is given that,
$v(t) = {t^3} - 4{t^2} - 3t + 2$ -----(1)
And we know that the rate of change of velocity is acceleration, so differentiation the given equation (1) we will get,
$a(t) = 3{t^2} - 8t - 3$ -----(2)
And as per the question, it is asked that what is the minimum acceleration attained, so we have to differentiate again, so differentiating the equation (2) we get,
$a'(t) = 6t - 8$
And now we have to check on what time this acceleration comes zero,
$0 = 6t - 8 \\\ \Rightarrow 6t = 8 \\\ \Rightarrow t = \dfrac{4}{3} \\\$
So, we can say that the obtained time is a minimum point of acceleration because $a'(t) < 0$ to the left of $t$ and $a'(t) > 0$ to the right of $t$.
Hence, the minimum obtained acceleration at time, $t = \dfrac{4}{3}$ is:
$a(t) = 3{t^2} - 8t - 3 \\\ \Rightarrow a\left( {\dfrac{4}{3}} \right) = 3{\left( {\dfrac{4}{3}} \right)^2} - 8\left( {\dfrac{4}{3}} \right) - 3 \\\ \Rightarrow a\left( {\dfrac{4}{3}} \right) = \dfrac{{16}}{3} - \dfrac{{32}}{3} - 3 \\\ \therefore a\left( {\dfrac{4}{3}} \right) = \dfrac{{ - 25}}{3} \\\$
Hence the minimum acceleration attained in the given interval of time is $\dfrac{{ - 25}}{3}$.

Note: Note that the obtained negative sign shows that the particle or object is not accelerating instead of that the particle is in retardation. We can also solve this problem if we have a good concept of parabolic equations as the equation.