A boy goes to market at an average speed $29 km h^{- 1}$ Finding the market closed, he immediately returns back by the same path with an average speed of $40 km h^{- 1}$ If the boy covers half the path during his return journey, find the average speed and average velocity.

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Solution

Step 1:Given data
The initial average speed of boy = $29 km h^{- 1}$
Return average speed of boy = $40 km h^{- 1}$
Let the distance covered by the boy to market be $s$
t₁ be the time taken to reach the market
t₂ be the time taken to return from the market.
Total timeT=t₁+t₂
distance covered by the boy in his return journey will be half of the initial value, ie; $\frac{s}{2}$
Step 2: Calculations
Average speed:
Average speed is defined as the ratio of total distance covered per total time taken.
Now, average speed = $\frac{total distance}{total time}$
Total distance = $S + \frac{S}{2}$
Rearranging,
Time taken= $\frac{d i s \tan c e}{s p e e d}$
$t_{1} = \frac{d i s \tan c e t o m a r k e t}{i n i t i a l s p e e d} = \frac{S}{29} - - - - (1)$
$t_{2} = \frac{D i s \tan c e f r o m m a r k e t t o h o m e}{r e t u r n a v e r a g e s p e e d} = \frac{\frac{S}{2}}{40} - - - (2)$
So,
Total time=Time taken to reach the market +Time taken to reach home from the market
$T o t a l t i m e T = t_{1} + t_{2} = \frac{S}{29} + \frac{\frac{s}{2}}{40} - - - - - (3)$

Average speed= $\frac{T o t a l d i s ta a n c e}{t o t a l t i m e} = \frac{S + \frac{S}{2}}{T} = \frac{S + \frac{S}{2}}{t_{1} + t_{2}}$
Substituting equations (1) and (2) in the place of t₁ and t₂
Average speed = $\frac{s + \frac{s}{2}}{\frac{s}{29} + \frac{\frac{s}{2}}{40}} = \frac{1.5 \times 80 \times 29}{109} = 31.93 km h^{- 1} .$
Average velocity:
Average velocity is defined as the ratio of the displacement to the total time taken.
Displacement is the shortest distance which is equal to $\frac{S}{2}$
$A v e r a g e v e l o c i t y = \frac{displacement}{total time} = \frac{\frac{S}{2}}{t_{1} + t_{2}} S u b s t i t u t i n g T h e v a l u e s o f t_{1} a n d t_{2} f r o m e q u a t i o n (1) a n d (2), A v e r a g e v e l o c i t y = \frac{s / 2}{\frac{s}{29} + \frac{s / 2}{40}} = \frac{29 \times 80}{2 \times 109} = 10.64 km h^{- 1} .$
Hence,
The average speed is 31.93kmh^-1 and the average velocity is 10.64kmh^-1

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A boy goes to market at an average speed 29kmh-1 Finding the market closed, he immediately returns back by the same path with an average speed of 40kmh-1 If the boy covers half the path during his return journey, find the average speed and average velocity.

A boy goes to market at an average speed $29 km h^{- 1}$ Finding the market closed, he immediately returns back by the same path with an average speed of $40 km h^{- 1}$ If the boy covers half the path during his return journey, find the average speed and average velocity.