# Question #b49a3

Jun 1, 2017

#### Explanation:

The formula is

$v \left(t\right) = {t}^{3} - 4 {t}^{2} + 4 t$

$\left(i\right)$,

After $2 s$

$v \left(2\right) = {2}^{3} - 4 \cdot {2}^{2} + 8 = 0 m {s}^{-} 1$

$\left(i i\right)$,

The derivative is

$\frac{\mathrm{dv}}{\mathrm{dt}} = v ' \left(t\right) = 3 {t}^{2} - 8 t + 4$

$\left(i i i\right)$

$\frac{\mathrm{dv}}{\mathrm{dt}} = 0$

When,

$3 {t}^{2} - 8 t + 4 = 0$

$\left(3 t - 2\right) \left(t - 2\right) = 0$, $\implies$

$t = \frac{2}{3}$, and $t = 2$

$\left(i v\right)$

Therefore

$v \left(\frac{2}{3}\right) = {\left(\frac{2}{3}\right)}^{3} - 4 \cdot {\left(\frac{2}{3}\right)}^{2} + 4 \cdot \frac{2}{3} = 1.19 m {s}^{-} 1$

$v \left(2\right) = {2}^{3} - 4 \cdot {2}^{2} + 4 \cdot 2 = 0$

The local maximum value is $= 1.19 m {s}^{-} 1$ and the minimum value is $= 0 m {s}^{1}$

Here is a graph
graph{x^3-4x^2+4x [0, 7.024, 0, 3.51]}