Question

A body travels a distance of $10\mathrm{km}$ with a constant speed of $30\mathrm{km}{\mathrm{h}}^{-1}$ and then the next $40\mathrm{km}$ at a constant speed of $50\mathrm{km}{\mathrm{h}}^{-1}$. Find the average speed for the whole journey.

Answer 1

Step1:
Given Data

Distance$\left({\mathrm{s}}_1\right)$ = $10\mathrm{km}$.

Initial speed$\left({\mathrm{v}}_1\right)$ = $30\mathrm{km}{\mathrm{h}}^{-1}$.

Final speed$\left({\mathrm{v}}_2\right)$ = $50\mathrm{km}{\mathrm{h}}^{-1}$.

distance$\left({\mathrm{s}}_2\right)$ = $40\mathrm{km}$.

Step2: To calculate - The average speed of the whole journey.

Now, average speed$\left(\mathrm{v}\right)$ = $\frac{totaldis\tan ce}{totaltime}$.

Total distance would be ${\mathrm{s}}_1+{\mathrm{s}}_2$

${\mathrm{s}}_1$ is covered with speed ${\mathrm{v}}_1$. Therefore, the time it will take to cover this distance would be ${\mathrm{t}}_1=\frac{{\mathrm{s}}_1}{{\mathrm{v}}_1}$.

${\mathrm{s}}_2$ is covered with speed ${\mathrm{v}}_2$. Therefore, the time it will take to cover this distance would be ${\mathrm{t}}_2=\frac{{\mathrm{s}}_2}{{\mathrm{v}}_2}$

$\mathrm{v}=\frac{{\mathrm{s}}_1+{\mathrm{s}}_2}{{\mathrm{t}}_1+{\mathrm{t}}_2}=\frac{{\mathrm{s}}_1+{\mathrm{s}}_2}{{\displaystyle \frac{{\mathrm{s}}_1}{{\mathrm{v}}_1}}+{\displaystyle \frac{{\mathrm{s}}_2}{{\mathrm{v}}_2}}}=\frac{10+40}{{\displaystyle \frac{10}{30}}+{\displaystyle \frac{40}{50}}}=\frac{50\times 15}{17}=44.12\mathrm{km}/\mathrm{h}$
Thus, the average speed for the whole journey is $\mathrm{v}=44.12\mathrm{km}/\mathrm{h}$