Question

A mining cart is pulled up a hill at 20 km/hand then pulled back down the hill at 35 km/hthrough its original level. (The time required for the cart’s reversal at the top of its climb is negligible).What is the average speed of the cart for its round trip, from its original level back to its original level?

Short answer

The average speed of the cart for its round trip is 25 km/h

Step-by-step solution

Given information

$$\begin{gathered} v_1=20\mathrm{km}/\mathrm{h} \\ {v}_2=35\mathrm{km}/\mathrm{h} \end{gathered}$$

To understand the concept

This problem involves simple physical quantities speed, distance, and acceleration. When acceleration is constant one can say that speed is the ratio of distance and time.

Calculations for the average speed of the cart

So, the total distance the cart is being pulled will be$2y\mathrm{km}$

Time for pulling the cart up is$t_1=\frac{y}{20}\mathrm{hrs}$

Time for bringing the cart down is$t_2=\frac{y}{35}\mathrm{hrs}$

So, the total time is

$$\begin{gathered} t=t_1+t_2 \\ =\frac{y}{20}+\frac{y}{35} \end{gathered}$$

The average velocity is defined as

localid="1660890442400" $$\begin{gathered} \mathrm{Average}\mathrm{speed}=\frac{\mathrm{Total}\mathrm{distance}}{\mathrm{Total}\mathrm{time}} \\ \mathrm{Average}\mathrm{speed}=\frac{2\mathrm{y}}{\left({\displaystyle \frac{\mathrm{y}}{20}+\frac{\mathrm{y}}{35}}\right)} \\ \mathrm{Average}\mathrm{speed}=\frac{2\mathrm{y}}{\left({\displaystyle \frac{55\mathrm{y}}{700}}\right)} \\ =25.45 \\ ~25\mathrm{km}/\mathrm{h} \end{gathered}$$

So, the average speed of the cart will be 25 km/h