# Question 3d192

Sep 5, 2016

Given a car is driven on a straight line. Its distance $x$ from a traffic light at red, starting point at $t = 0$; is given as a function of time $t$
$x \left(t\right) = 2 {t}^{2} + 0.25 {t}^{3}$

First step: lets find $x \left(0\right) , x \left(2\right) \mathmr{and} x \left(4\right)$
$x \left(0\right) = 2 \times {0}^{2} + 0.25 \times {0}^{3} = 0 m$
$x \left(2\right) = 2 \times {2}^{2} + 0.25 \times {2}^{3} = 8 + 2 = 10 m$
$x \left(4\right) = 2 \times {4}^{2} + 0.25 \times {4}^{3} = 32 + 16 = 48 m$
We know that "Average Velocity"=("Displacement " Delta x)/("Time of travel "Delta t)#
1. $t = 0 s$ to $t = 2 s$
$\text{Average Velocity} = \frac{x \left(2\right) - x \left(0\right)}{2 - 0} = \frac{10 - 0}{2} = 5 m {s}^{-} 1$
2. $t = 2 s$ to $t = 4 s$
$\text{Average Velocity} = \frac{x \left(4\right) - x \left(2\right)}{4 - 2} = \frac{48 - 10}{2} = 19 m {s}^{-} 1$
3. $t = 0 s$ to $t = 4 s$
$\text{Average Velocity} = \frac{x \left(4\right) - x \left(0\right)}{4 - 0} = \frac{48 - 0}{4} = 12 m {s}^{-} 1$