# Question #5ef69

Jun 2, 2017

#### Explanation:

$\left(v\right)$

$\frac{\mathrm{dv}}{\mathrm{dt}}$ is the acceleration of the kite

$\left(v i\right)$
graph{x^3-4x^2+4x [-0.845, 10.25, -1.685, 3.865]}

The gradients at $t = 0.7$ and $t = 2$ is $= 0$

The acceleration is $= 0$ at $t = 0.7$ and $t = 2$

This is the local max. and the local min.

$\left(v i i\right)$

The area under the curve represents the distance travelled.

You can estimate the distance travelled by the kite by counting the number of squares between the curve and the x-axis

$\left(v i i\right)$

The distance is

$s = \int v \left(t\right)$dt

$= {\int}_{0}^{4} \left({t}^{3} - 4 {t}^{2} + 4 t\right) \mathrm{dt}$

$= {\left[{t}^{4} / 4 - \frac{4}{3} {t}^{3} + \frac{4}{2} {t}^{2}\right]}_{0}^{4}$

$= \left(4 \cdot 4 \cdot 4 - \frac{256}{3} + 32\right) - \left(0 + 0 + 0\right)$

$= 64 + 32 - \frac{256}{3}$

$= 96 - \frac{256}{3}$

$= \frac{32}{3}$

$= 10.67$

I hope that this is helpful !!