Question

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?​

Answer 1

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Here we go ~

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

So, time taken in this span is :

$\qquad \sf  \dashrightarrow \: \dfrac{40}{30}$

$\qquad \sf  \dashrightarrow \: \dfrac{4}{3} \: \:hours$

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 :

$\qquad \sf  \dashrightarrow \: \dfrac{40}{x} \: \: kmph$

Now, we know the formula to find average velocity ~

$\qquad \sf  \dashrightarrow \: v_{avg} = \dfrac{total \: \: distance \: \: covered}{total \: \: time \: \: taken}$

$\qquad \sf  \dashrightarrow \: 40 = \dfrac{80}{ \frac{40}{30} + \frac{40}{x} }$

$\qquad \sf  \dashrightarrow \: \dfrac{40}{30} + \dfrac{4 0}{x} = \dfrac{80}{40}$

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

$\qquad \sf  \dashrightarrow \: \dfrac{1}{30} + \dfrac{1}{ {x}^{} } = \dfrac{2}{40}$

$\qquad \sf  \dashrightarrow \: \dfrac{1}{x} = \dfrac{1}{20} - \dfrac{1}{30}$

$\qquad \sf  \dashrightarrow \: \dfrac{1}{x} = \dfrac{3 - 2}{60}$

$\qquad \sf  \dashrightarrow \: \dfrac{1}{x} = \dfrac{1}{60}$

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

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 ~