Question

If a particle moves along a straight line according to $s\left( t \right) = {t^4} - 4{t^3} + 6{t^2} - 20$, how do you find the maximum & minimum acceleration on $0 \leqslant t \leqslant 3$?

Answer 1

The question has given an equation of displacement with respect to time. We have to find the second derivative in order to find the acceleration. Then by substituting different values of time to the equation of acceleration we will find the maximum and minimum value of it.

Complete step by step solution:
It is given in the question that a particle moves along a straight line with the relation $s\left( t \right) = {t^4} - 4{t^3} + 6{t^2} - 20$. We have to find the maximum & minimum acceleration on $0 \leqslant t \leqslant 3$.
The relation given here is the change of displacement of the particle with time.
The first derivative of displacement with time will give us the velocity.
Hence, the first derivative of the given equation with respect to time is,
$\dfrac{d}{{dt}}\left( s \right) = 4{t^3} - 12{t^2} + 12t$
Now, the second derivative of displacement with time gives us acceleration.
Hence, the second derivative of the given equation is,
$\dfrac{{{d^2}}}{{d{t^2}}}\left( s \right) = 12{t^2} - 24t + 12$
It is the equation of the acceleration of the particle.
Let the acceleration of the particle be $a = 12{t^2} - 24t + 12$.
It is clear that the acceleration of the particle is dependent on time.
In between $0 \leqslant t \leqslant 3$, let us find the maximum and minimum acceleration by substituting the different values of time $t$ between $0$ and $3$.
Let us consider when time $t = 0$ then,
$a = 12 \times 0 - 24 \times 0 + 12$
$\Rightarrow a = 12{\text{ }}units$
When time $t = 1$ then,
$a = 12 \times {1^2} - 24 \times 1 + 12$
$\Rightarrow a = 0{\text{ }}units$
When time $t = 2$ then,
$a = 12 \times {2^2} - 24 \times 2 + 12$
$\Rightarrow a = 12{\text{ }}units$
When time $t = 3$ then,
$a = 12 \times {3^2} - 24 \times 3 + 12$
$\Rightarrow a = 48{\text{ }}units$
Hence it is clear that the acceleration is minimum at $t = 1$, $a = 0{\text{ }}units$ and maximum at $t = 3$, $a = 48{\text{ }}units$.

Note:
It must be noted that the first derivative of displacement with respect to time is velocity and the second derivative of displacement with respect to time is acceleration. Since the equation has no assigned unit for the given question, we also must not mention any particular unit in the answer.