Answer
a) Its acceleration is not constant.
b)
i) $a_{av} = 12.77 m/s^{2}$
ii) $a_{av} = 3.50 m/s^{2}$
iii) $a_{av} = 0.81 m/s^{2}$
Work Step by Step
a) $a(t) = v'(t)$. If $a$ was constant, the funtion $v(t)$ would be linear. It is not the case, so the acceleration is not constant.
b) Let's convert the velocities [$mi/h$] to velocities [$m/s$] :
$v(t = 0) = 0 mi/h = 0 m/s$
$v(t = 2.1) = 60 mi/h = 26.82 m/s$
$v(t = 20.0) = 200 mi/h = 89.41 m/s$
$v(t = 53) = 260 mi/h = 116.23 m/s$
i) $a_{av} = \dfrac{\Delta v}{\Delta t} = \dfrac{v(2.1)-v(0)}{2.1 - 0} = \dfrac{26.82 - 0}{2.1 - 0} = 12.77 m/s^{2}$
ii) $a_{av} = \dfrac{\Delta v}{\Delta t} = \dfrac{v(20.0)-v(2.1)}{20.0 - 2.1} = \dfrac{89.41 - 26.82}{20.0 - 2.1} = 3.50 m/s^{2}$
iii) $a_{av} = \dfrac{\Delta v}{\Delta t} = \dfrac{v(53)-v(20.0)}{53 - 20.0} = \dfrac{116.23-89.41}{53 - 20.0} = 0.81 m/s^{2}$
On the graph, we see that the velocity is less and less growing, which means that its derivative (the acceleration) is decreasing. These results also show that the acceleration is decreasing.
