A person travels the total distance in two parts in the ratio $2 : 1$ with a constant speed of $30 k m h^{- 1}$ in the first part and $40 k m h^{- 1}$ in the second part. What is the average speed of the journey?

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Solution

Let total distance be $x$
First Part
Distance, $d_{1} = \frac{2 x}{3}$
Time, $t_{1} = \frac{d_{1}}{v_{1}} = \frac{2 x}{3 \times 30} = \frac{2 x}{90}$
Second Part
Distance, $d_{2} = \frac{x}{3}$
Time, $t_{2} = \frac{d_{2}}{v_{2}} = \frac{x}{3 \times 40} = \frac{x}{120}$
Average Speed
$v_{a v} = \frac{T o t a l d i s t a n c e}{T o t a l t i m e}$
$= \frac{x}{\frac{2 x}{90} + \frac{x}{120}}$
$= \frac{1}{\frac{1}{45} + \frac{1}{120}}$
$= \frac{1}{\frac{8}{360} + \frac{3}{360}}$
$= \frac{360}{11}$
$= 32.7 k m / h$

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A person travels the total distance in two parts in the ratio 2:1 with a constant speed of 30kmh−1 in the first part and 40kmh−1 in the second part. What is the average speed of the journey?

Let total distance be xFirst PartDistance, d1=2x3Time, t1=d1v1=2x3×30=2x90Second PartDistance, d2=x3Time, t2=d2v2=x3×40=x120Average Speedvav=TotaldistanceTotaltime =x2x90+x120 =1145+1120 =18360+3360 =36011 =32.7km/h

A person travels the total distance in two parts in the ratio $2 : 1$ with a constant speed of $30 k m h^{- 1}$ in the first part and $40 k m h^{- 1}$ in the second part. What is the average speed of the journey?