Question

A Train travels the first 15 km at a uniform speed of 30 km/h , the next 75 km at a uniform speed of 50 km/h and the last 10 km at a uniform speed 20 km/h. Calculate the average speed for entire train journey.

Answer 1

Step 1: Given data and to find

$$\begin{gathered} {\mathrm{u}}_1=30{\mathrm{kmh}}^{-1}\mathrm{at}\mathrm{d}=15\mathrm{km} \\ {\mathrm{u}}_2=50{\mathrm{kmh}}^{-1}\mathrm{at}\mathrm{d}=75\mathrm{km} \\ {\mathrm{u}}_3=20{\mathrm{kmh}}^{-1}\mathrm{at}\mathrm{d}=10\mathrm{km} \end{gathered}$$

Step 2: Definition

1. Speed is defined as the distance travelled by the body at a given time.
2. Average speed is defined as the ratio of total distance travelled by the total time taken by a body at different instances.

Step 3: Formulas used

$$\begin{gathered} \mathrm{Speed}=\frac{\mathrm{distance}}{\mathrm{time}} \\ \mathrm{Average}\mathrm{speed}=\frac{\mathrm{total}\mathrm{distance}\mathrm{covered}}{\mathrm{total}\mathrm{time}\mathrm{taken}} \end{gathered}$$

Step 4: Calculating the speed when $d=15kmand{u}_1=30km{h}^{-1}$

$$\begin{gathered} \mathrm{speed}=\frac{\mathrm{distance}}{\mathrm{time}} \\ \mathrm{time}=\frac{\mathrm{distance}}{\mathrm{speed}} \\ {\mathrm{t}}_1=\frac{15}{30}=\frac{1}{2}\mathrm{h} \end{gathered}$$

Step 5: Calculating the speed when $d=75kmand{u}_2=50km{h}^{-1}$

$$\begin{gathered} \mathrm{speed}=\frac{\mathrm{distance}}{\mathrm{time}} \\ \mathrm{time}=\frac{\mathrm{distance}}{\mathrm{speed}} \\ {\mathrm{t}}_2=\frac{75}{50}=\frac{3}{2}\mathrm{h} \end{gathered}$$

Step 6: Calculating the speed when$d=10kmand{u}_3=20km{h}^{-1}$

$$\begin{gathered} \mathrm{speed}=\frac{\mathrm{distance}}{\mathrm{time}} \\ \mathrm{time}=\frac{\mathrm{distance}}{\mathrm{speed}} \\ {\mathrm{t}}_3=\frac{10}{20}=\frac{1}{2}\mathrm{h} \end{gathered}$$

Step 7: Calculating average speed

$$\begin{gathered} \mathrm{Average}\mathrm{speed}=\frac{\mathrm{total}\mathrm{distance}\mathrm{travelled}}{\mathrm{total}\mathrm{time}\mathrm{taken}} \\ \mathrm{total}\mathrm{distance}\mathrm{covered}=15+75+10=100\mathrm{km} \\ \mathrm{total}\mathrm{time}\mathrm{taken}=\frac{1}{2}+\frac{3}{2}+\frac{1}{2}=\frac{5}{2}\mathrm{h} \\ \mathrm{Therefore},\mathrm{average}\mathrm{speed}=\frac{100}{{\displaystyle \frac{5}{2}}}=\frac{200}{5}=40{\mathrm{kmh}}^{-1} \end{gathered}$$

Therefore, the average speed for the entire train journey $40km{h}^{-1}$.