A cyclist goes uphill at a speed of 8 km/h and downhill at a speed of 32 km/h. If uphill and downhill journeys involve the same distance, what is the average speed during the whole journey?

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Solution

Step 1: Given data
The uphill speed $v_{1} = 8 k m / h$ .
The downhill speed $v_{2} = 32 k m / h$ .
The uphill and downhill journeys involve the same distance.
Now, we know that the average speed is the ratio of total distance traveled by the body and total time taken by the body to cover the total distance.
The formula for calculating the average velocity is $A v e r a g e s p e e d (v) = \frac{T o t a l d i s \tan c e t r a v e l e d}{T o t a l t i m e t a k e n}$ .
$T i m e = \frac{T o t a l d i s \tan c e t r a v e l e d}{A v e r a g e s p e e d}$
Now, let $d_{1}$ be the distance for uphill and $d_{2}$ be the distance for downhill.
As in the question it is given to us that the uphill and downhill journeys involve the same distance (say d)
We can write;
$d = d_{1} = d_{2}$
Step 2: Finding time intervals for the uphill and downhill journeys
Let t₁ be the time for the uphill journey and t₂ for the downhill journey.
So by using the formula, we can write, $t_{1} = \frac{d}{v_{1}}$ and $t_{2} = \frac{d}{v_{2}}$ .
Step 3: Finding the value of average velocity
$v_{a v g} = \frac{t o t a l d i s \tan c e}{t o t a l t i m e} = \frac{d_{1} + d_{2}}{t_{1} + t_{2}}$
Putting the values of t₁, t₂, and total distance.
$v_{a v g} = \frac{2 d}{\frac{d}{v_{1}} + \frac{d}{v_{2}}} = \frac{2 d}{\frac{d (v_{2} + v_{1})}{v_{1} v_{2}}} = \frac{2 \times v_{1} \times v_{2}}{v_{1} + v_{2}} = \frac{2 \times 8 \times 32}{8 + 32} = \frac{16 \times 32}{40} = \frac{4 \times 32}{10} = 12.8 k m / h .$
Hence, the average speed will be 12.8 km/h.

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