A bus leaving a bus stop accelerates at 0.8 $ms^{-2}$ for 5s and then travels at a constant speed for 2 minutes before slowing down uniformly at 4 $ms^{-2}$ to come to rest at the next bus stop. Calculate

i) the constant speed
ii) the distance travelled while the bus is accelerating
iii) the total distance travelled.

Question

A bus leaving a bus stop accelerates at 0.8 $ms^{-2}$ for 5s and then travels at a constant speed for 2 minutes before slowing down uniformly at 4 $ms^{-2}$ to come to rest at the next bus stop. Calculate

i) the constant speed
ii) the distance travelled while the bus is accelerating
iii) the total distance travelled.

Accepted Answer

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

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

[latex]\begin{matrix} \overset{\boxed{ ext{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 }\ { ext{ii) Let the distance travelled be }} {s_{1}} m\end{matrix}[/latex].

[latex]\begin{matrix}u = 0, a = 0.8, t = 5, ext{ so use } s = ut + \frac{1}{2}at^{2} &\longleftarrow \boxed{\begin{matrix} ext{Want s } \ ext{ know } u = 0, t = 5, a = 0.8 \ s = \frac{1}{2}(u + v)t ✘ \ s = ut + \frac{1}{2}at^{2} ✓ \end{matrix}}\end{matrix}[/latex]

[latex]s_{1} = 0 + \frac{1}{2} × 0.8 × 5²[/latex]

= 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 [latex]s_{3}[/latex] m.

[latex]\begin{matrix} \overset{ \quad\quad\quad \quad\boxed{\begin{matrix} ext{ Want s}\ ext{ 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, ext{so use } v^{2} = u^{2} } {+} 2as\end{matrix}[/latex]

[latex]0 = 16 + 2(-0.4)s_{3}[/latex]

0.8[latex]s_{3}[/latex] = 16

[latex]s_{3} = 20[/latex]

Distance taken to slow down = 20 m.

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

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

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