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On a 80 km track, a train travels 40 km with a uniform speed of 30 km h¹. How fast must the train travel the next 40 km so as to have average speed 40 km h-¹ for the entire trip?​

User Ralokt
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{ \qquad\qquad\huge\underline{{\sf Answer}}}

Here we go ~

In first 40 km, it travels with speed of 30 kmph

So, time taken in this span is :


\qquad \sf  \dashrightarrow \: (40)/(30)


\qquad \sf  \dashrightarrow \: (4)/(3) \: \:hours

And let's assume that it travels with uniform speed of x kmph during next 40 km

Now, time taken in this span of time is :


\qquad \sf  \dashrightarrow \: (40)/(x) \: \: kmph

Now, we know the formula to find average velocity ~


\qquad \sf  \dashrightarrow \: v_(avg) = (total \: \: distance \: \: covered)/(total \: \: time \: \: taken)


\qquad \sf  \dashrightarrow \: 40 = (80)/( (40)/(30) + (40)/(x) )


\qquad \sf  \dashrightarrow \: (40)/(30) + (4 0)/(x) = (80)/(40)


\qquad \sf  \dashrightarrow \: 40 \bigg( (1)/(30) + \frac{1}{ {x}^{} } \bigg) = (80)/(40)


\qquad \sf  \dashrightarrow \: (1)/(30) + \frac{1}{ {x}^{} } = (2)/(40)


\qquad \sf  \dashrightarrow \: (1)/(x) = (1)/(20) - (1)/(30)


\qquad \sf  \dashrightarrow \: (1)/(x) = (3 - 2)/(60)


\qquad \sf  \dashrightarrow \: (1)/(x) = (1)/(60)


\qquad \sf  \dashrightarrow \: x = 60 \: \: kmph

So, for next 40 km, he need to maintain average speed of 60 kmph, in order to keep an average of 40 kmph for the whole trip ~

User Bernard Leech
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