Petrol stop

Answer: 1 hour and 10 minutes

Using a distance-time graph

Image

Gradient represents speed so gradients are $80$ and $100$

So vertical distances are $80T_1$ and $100T_2$

time: $T_{1}+T_{2}=3$ hours$-20$ minutes $=2\frac23$ hours

distance: $80T_{1}+100T_{2}=250\Rightarrow 8T_1+10T_2=25$

$T_1+T_2 = 2\frac23$    $\times8$ gives $8T_1+8T_2=16\tfrac{16}3=21\frac13$

Subtracting, $$\begin{array}{rl}(8T_1+10T_2)-(8T_1+8T_2)& =25-21\frac{1}{3}\\ 2T_2& =3\frac{2}{3}=\frac{11}{3}\\ T_2& =\frac{11}{6}\end{array}$$ $\frac{11}6$ hours $=1$ hour and $50$ minutes.

Using a speed-time graph

Image

time: $T_{1}+T_{2}=3$ hours$-20$ minutes $=2\frac23$ hours

Distance on a speed-time graph is given by the area under the graph, so: $$\begin{array}{rl}80T_1+100T_2& =250\\ \Rightarrow 80(T_1+T_2)+20T_2& =250\\ \Rightarrow 80\times 2\frac{2}{3}+20T_2& =250\\ \Rightarrow 8\times \frac{8}{3}+2T_2& =25\\ \Rightarrow 2T_2& =\frac{75}{3}-\frac{64}{3}\\ \Rightarrow T_2& =\frac{11}{6}\end{array}$$ $\frac{11}6$ hours $=1$ hour and $50$ minutes.

Using the speed-distance-time relationship

Distance = speed $\times$ time

Distance travelled before the petrol stop is $80\times T_{1}$

Distance travelled after the petrol stop is $100\times T_{2}$

Total distance is $250$ km so $80T_{1}+100T_{2}=250$.

Total time is $3$ hours. $20$ minutes $=\frac13$ hour, so $T_{1}+T_{2}=2\frac23$.

$T_{1}+T_{2}=2\frac23\Rightarrow T_1 = \frac83 - T_2$

Substitute into other equation: $$\begin{array}{rl}80(\frac{8}{3}-T_2)+100T_2& =250\\ \Rightarrow \frac{64}{3}-8T_2+10T_2& =25\\ \Rightarrow \frac{64}{3}+2T_2& =\frac{75}{3}\\ \Rightarrow T_2& =\frac{11}{6}\end{array}$$ $\frac{11}6$ hours $=1$ hour and $50$ minutes.