#### Answer

a) The average acceleration of the car between and $t = 5$s is $a_{x} = 0.5$m s$^{-1}$.
b) The instantaneous acceleration is $a_{x} = 0$m s$^{-1}$ at $t = 0$s, and $a_{x} = 1$m s$^{-1}$ at $t = 5$s.
c) The answer is provided as an image.

#### Work Step by Step

a) The car's velocity as a function of time is given as:
$v_{x}(t) = \alpha + \beta t^{2}$ where $\alpha = 3$m s$^{-1}$ and $\beta = 0.1$m s$^{-3}$.
To find the average acceleration between $t = 0$s and $t = 5$s the relationship between acceleration, velocity and time can be used:
$a_{x-av} = \frac{\Delta v_{x}}{\Delta t}$
First find $\Delta v_{x}$:
$v_{x}(t) = 3 + 0.1 t^{2}$
at $t = 0$s
$v_{1x} = 3 + 0.1\times 0^{2}$
$v_{1x} = 3$m s$^{-1}$.
at $t = 5$s
$v_{2x} = 3 + 0.1\times 5^{2}$
$v_{2x} = 3 + 0.1\times 25$
$v_{2x} = 3 + 2.5$
$v_{2x} = 5.5$m s$^{-1}$.
$\Delta v = v_{2x} - v_{1x} = 5.5 - 3 = 2.5$m s$^{-1}$.
$\Delta t = 5 - 0 = 5$s
Now acceleration can be found:
$a_{x-av} = \frac{2.5}{5} = 0.5$m s$^{-2}$
b) Find the instantaneous acceleration at $t = 0$s and $t = 5$s.
Acceleration is the rate of change of velocity with respect to time. This means acceleration is the first derivative of velocity, or the second derivative of displacement. An equation for acceleration can be calculated through differentiation:
$a_{x} = \frac{d v_{x}}{dt}$
$v_{x} = 3 + 0.1\times t^{2}$
$a_{x} = 2\times 0.1 t^{2-1}$
$a_{x} = 0.2 t$
at $t = 0$s
$a_{x} = 0.2\times 0$
$a_{x} = 0$ m s$^{-2}$.
at $t = 5$s
$a_{x} = 0.2\times 5$
$a_{x} = 1$ m s$^{-2}$.
c) To graph the acceleration - time and velocity - time graphs, we must examine the equations to find the key points.
$v_{x} = 3 + 0.1 t^{2}$ and $a_{x} = 0.2 t$.
The relationship between velocity and time is parabolic, and has a y-intercept (initial $v_{x}$ value of 3m/s at t = 0 s), and a final velocity of 5.5m/s at t = 5 s ( as calculated in part a).
Using these key points, the graph can be sketched. Label the axis accordingly with units, and title the graph.
The acceleration-time relationship is linear, with a gradient of 0.2. As found in part b), the initial acceleration (y-intercept, at t = 0) is 0m s$^{-2}$, and the final acceleration (t = 5) is 1 m s$^{-2}$. This can be sketched on the graph. Don't forget axis titles, units and a graph title!