Average speed of train

Question

Distance between two stations $A$ and $B$ is $778$km. A train covers the journey from $A$ to $B$ at a uniform speed of $84$km per hour and returns back to $A$ with a uniform speed of $56$km per hour. Find the average speed of the train during the whole journey?

The correct answer is:

Let distance between $A$ and $B$ be $x$
Time taken for travelling from $A$ to $B$ is $\frac{x}{84}$
Time taken for travelling from $B$ to $A$ is $\frac{x}{56}$
Total distance travelled is $x+x=2x$
Total time taken is $\frac{x}{84}+\frac{x}{56}$
Average speed is $67.2$

I know speed is distace/time and I dont know if it may sound stupid but what i thought of was

$$avg=\frac{\text{speed}_1+\text{speed}_2}{2} \tag{This is how we calculate average}$$ $$\frac{84+56}{2}=70$$
why is it giving wrong answer?
Also do I need to revise my physics concepts or maths or both?

Answer 1

The formula for the average speed is: $Average Speed= \frac{Total Distance}{Total Time}$

To obtain the total distance you simply add the distances. $778+778$. Then, you must obtain the total time.

That's where the "Time taken for travelling from B to A" in your solution comes in. Since you have the average speed in kilometers per hour, if you divide the total speed by the average speed, you will find out how many hours it took to travel that distance, so:

$\frac{778}{56} = 13.89$ and $\frac{778}{84}=9.26$

Adding the total hours and using the average speed formula, you obtain: $Average Speed = \frac{1556}{23.15}=67.2$

In your answer you simply took the average of the speed averages, but not the average speed for the journey.

So, you added $56=\frac{778}{13.89}$ and $84=\frac{778}{9.26}$

$\frac{778}{13.89}+\frac{778}{9.26} = \frac{778*13.89 + 778*9.26}{13.89*9.26} = \frac{18010.7}{128.6214} = 140$

Then, dividing it by $2$, you obtained 70.

What you should have done was simply adding: $\frac{778+778}{13.89+9.26} = \frac{1556}{23.15} = 67.2$

Answer 2

An average speed is a ratio of a distance travelled to a travelling time.

An average distance in the first pass is $v_1 = s/t_1$ and an average speed in the returning pass on the same distance is $v_2 = s/t_2$.

The average speed on the whole travel is a total distance to total time used: $$v=\frac{s+s}{t_1+t_2}$$ which results in: $$v = \frac 2{\frac{t_1+t_2}s} =\frac 2{\frac {t_1} s+\frac {t_2} s}=\frac 2{\frac 1{v_1}+\frac 1{v_2}}$$