Question 1.4: A bus leaving a bus stop accelerates at 0.8 ms^-2 for 5s and...

i) The diagram shows the information for the first part of the motion.
Let the constant speed be v $ms^{-1}$.

$\begin{matrix}u = 0, \ a = 0.8, \ t = 5,\ \text{so use } v = u + at &\longleftarrow \\ \end{matrix} \boxed{\begin{matrix} \text{Want v}\\ \text{ Know }u = 0,\ t = 5 ,\ a = 0.8\ \\v^{2} = u^{2} + 2as      ✘         v = u + at        ✓  \end{matrix}}$
v = 0 + 0.8 × 5
= 4
The constant speed is 4 $ms^{-1}$.

$\begin{matrix} \overset{\boxed{\text{Use the suffix because there are three distances to be found in this question.}}}{\quad \quad \quad \quad \quad\quad \quad \quad \quad \quad\quad \quad \quad \downarrow }\\ {\text{ii) Let the distance travelled be }} {s_{1}} m\end{matrix}$.

$s_{1} = 0 + \frac{1}{2} × 0.8 × 5²$

= 10

The bus accelerates over 10 m.

iii) The diagram gives all the information for the rest of the journey.

Between B and C the velocity is constant so the distance travelled is 4 × 120 = 480 m.

Let the distance between C and D be $s_{3}$ m.

$\begin{matrix} \overset{ \quad\quad\quad \quad\boxed{\begin{matrix}\text{ Want s}\\\text{ know u = 4, a = – 0.4, v = 0 }\\ v = u + at  ✘ \\ s = ut + \frac{1}{2}at^{2}  ✘ \\ s = \frac{1}{2}(u + v)t  ✘\\ v^{2} = u^{2} + 2as  ✓\end{matrix}}}{\quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad \quad \quad \quad \quad \downarrow }\\ {u = 4, a = -0.4, v = 0, \text{so use } v^{2} = u^{2} } {+} 2as\end{matrix}$

$0 = 16 + 2(-0.4)s_{3}$

0.8$s_{3}$ = 16

$s_{3} = 20$

Distance taken to slow down = 20 m.

The total distance travelled is (10 + 480 + 20) m = 510 m.