\(41.7 km / hr\)
\(10 min =10 / 60 hr\)
for the first journey
speed \(=\) distance/time
distance \(1=\) speed \(\times\) time
$$
\begin{aligned}
&=50 km / hr \times \frac{10}{60} hr \\
&=8.33 km
\end{aligned}
$$or the second journeyistance \(2=\) speed \(\times\) time
\(=\frac{25 km }{ hr } \times \frac{5}{60} hr =2.083 km\)
Total distance \(=8.33+2.083\)
Total time \(=10+5=15.413 km\)
\(5=15 min\)
\(=(10 / 60) hr\)
\(=\frac{\left(\begin{array}{c}\text { total distance } \\ \text { travelled }\end{array}\right)}{\text { total time taken }}\)
\(=\frac{10.413 km }{\frac{15}{60} hr }=\frac{10.413 \times 60}{15}\)
\(=41.65 km / hr \approx 41.7 km / hr\)
Note: We did not recognize the given angles because we are only asked to calculate the average speed which is a scalar quantity, if however we are asked to calculate the average velocity (a vector quantity) than we recognize the given angles as illustrated below
let the resultant displacement be \(R\)
disp. \(A B=\frac{50 km }{ hr } \times \frac{10}{60}=8.33 km\) disp. \(B C=\frac{25 km }{ hr } \times \frac{5}{60}=2.083 km\)
\(\therefore\) by cosine rule
$$
\begin{aligned}
&R^{2}=8.33^{2}+2.083^{2}-2(8.33)(2.083) \cos \left(135^{\circ}\right) \\
&R^{2}=69.389+4.339-[34.703(-0.7071)] \\
&R^{2}=73.728+24.538 \\
&R^{2}=98.266 \\
&R=\sqrt{98.266}
\end{aligned}
$$
\(R \approx 9.91 km\)
Hence average velocity
\(=\) total or resultant \(=\frac{\text { displacement }}{\text { time taken }}\)
\(=\frac{9.91 km }{\frac{(10+5) hr }{60}}=\frac{9.91}{\frac{15}{60}}=\frac{9.91 \times 60}{15}\)