Question

Rohan, while driving to school computes the average speed for his trip to be 30
km/h. On his return trip along the same route, there is less traffic and average speed
is 50 km/h. What is the average speed of rohan's trip

Answer 1

$\huge \pink{ \boxed{ \sf{ \green{Answer:-}}}}$

As per given data in the question:

Average speed from home to school v1av = 30km h-1

Average speed from School to Home v2av = 50km h-1

Our assumptions:

The distance Rohan commutes while driving from Home to School = S

Let us assume time taken by Rohan to commutes this distance = t1

Distance Rohan commutes while driving from School to Home = S

Let us assume time taken by Rohan to commutes this distance = t2

Also we know that:

The time taken from Home to School t1

$= \frac{s}{v1av}$

Similarly Time taken from School to Home t2

$= \frac{s}{v2av}$

Total distance:

Total distance from home to school and backward

= 2S——-(i)

Total time taken from home to school and backward (T)

= $\frac{s}{30} + \frac{s}{50}$ ———-(ii)

Formula:

Speed = Distance / Time

Therefore,

Average Speed(Vav)for covering total distance (2s)

$= \frac{Total \: Distance}{Total \: Time}$

So from equations (i) and (ii) we get,

Average Speed(Vav)

$\huge = \frac{2s}{( \frac{s}{30} + \frac{s}{50} )}$

$\huge = \frac{2s}{( \frac{3s + 5s}{150} )}$

$\huge = \frac{2s}{8s} \times 150$

$\huge = 37.5kmh - 1$

Final Answer:

The average speed for Rohan’s trip = 37.5kmh-1

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