What is the average speed for the total journey? Full question the description box below as an image.

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Answer 1 · 2018-01-04T06:41:03

The answer is $S = \frac{160 x}{80 + x}$ $S=(160x)/(80+x)$

Let the total distance of the journey be $= d k m$ $=d km$

Haf the distance is $= \frac{d}{2} k m$ $=d/2km$

Average speed for the first half is $= 80 k m h^{- 1}$ $=80kmh^-1$

The time taken for this first half is $t_{1} = \frac{\frac{d}{2}}{80} = (\frac{d}{160}) h$ $t_1=(d/2)/(80)=(d/160) h$

Average speed for the second half is $= x k m h^{- 1}$ $=xkmh^-1$

The time taken for this second half is $t_{2} = \frac{\frac{d}{2}}{x} = (\frac{d}{2 x}) h$ $t_2=(d/2)/(x)=(d/(2x)) h$

The total time is $t = t_{1} + t_{2} = \frac{d}{160} + \frac{d}{2 x} = d (\frac{1}{160} + \frac{1}{2 x})$ $t=t_1+t_2=d/160+d/(2x)=d(1/160+1/(2x))$

$= \frac{d}{2} (\frac{1}{80} + \frac{1}{x}) = \frac{d}{2} (\frac{80 + x}{80 x})$ $=d/2(1/80+1/x)=d/2((80+x)/(80x))$

The $average speed = \frac{total distance}{total time}$ $"average speed"="total distance"/"total time"$

$¯ v = \frac{d}{\frac{d}{2} (\frac{80 + x}{80 x})} = \frac{160 x}{80 + x}$ $barv=(d)/(d/2((80+x)/(80x)))=(160x)/(80+x)$

$S = \frac{160 x}{80 + x}$ $S=(160x)/(80+x)$