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Question: The distance between Delhi and Agra is \(200km\). A train travels the first \(100km\) at a speed of ...

The distance between Delhi and Agra is 200km200km. A train travels the first 100km100km at a speed of 50km/h50km/h. How fast must the train travel the next 100km100km, so as to average 70km/h70km/h for the whole journey?

Explanation

Solution

We can calculate the time to cover the first half of the journey, and the time to cover the second half can be obtained in terms of the unknown speed. The total time will be the sum of the two times. Dividing the total distance by the total time, we will get the average speed with which we can obtain an equation in terms of the unknown speed solving which we will get the final answer.

Complete step-by-step solution:
Let the next 100km100km be travelled by the train with a speed of v km/hv{\text{ k}}m/h
We know that the average velocity is defined as the total distance covered by an object divided by the total time taken. According to the question, the train travels between Delhi and Agra which are separated by a distance of 200km200km. So the total distance covered by the train is equal to 200km200km, that is we have
d=200kmd = 200km...............(1)
Now, the journey of the train is split into two equal halves of the total distance to be covered. The first half distance of 100km100km is covered with a speed of 50km/h50km/h. So the time taken to complete the first half of the journey is given by
t1=100km50km/h{t_1} = \dfrac{{100km}}{{50km/h}}
t1=2h\Rightarrow {t_1} = 2h............(2)
Also, according to our assumption, the next half distance of is covered with a speed of . So the time taken to complete the second half of the journey is given by
t2=100kmv km/h{t_2} = \dfrac{{100km}}{{v{\text{ }}km/h}}
t2=100vh\Rightarrow {t_2} = \dfrac{{100}}{v}h ……………...(3)
So the total time of the complete journey is given by
t=t1+t2t = {t_1} + {t_2}
Putting (2) and (3) in the above equation, we get
t=2h+100vht = 2h + \dfrac{{100}}{v}h
t=(2+100v)h\Rightarrow t = \left( {2 + \dfrac{{100}}{v}} \right)h..............(4)
Now, according to the definition of the average speed, it is given by
vavg=dt{v_{avg}} = \dfrac{d}{t}
Putting (1) and (4) in the above equation, we get
vavg=200km(2+100v)h{v_{avg}} = \dfrac{{200km}}{{\left( {2 + \dfrac{{100}}{v}} \right)h}}
vavg=200(2+100v)km/h\Rightarrow {v_{avg}} = \dfrac{{200}}{{\left( {2 + \dfrac{{100}}{v}} \right)}}km/h
Now, according to the question, the average speed of the train for the whole journey is equal to 70km/h70km/h. Therefore substituting vavg=70km/h{v_{avg}} = 70km/h in the above equation, we have
70=200(2+100v)70 = \dfrac{{200}}{{\left( {2 + \dfrac{{100}}{v}} \right)}}
2+100v=20070\Rightarrow 2 + \dfrac{{100}}{v} = \dfrac{{200}}{{70}}
Subtracting 22 from both the sides, we have
100v=200702\dfrac{{100}}{v} = \dfrac{{200}}{{70}} - 2
100v=67\Rightarrow \dfrac{{100}}{v} = \dfrac{6}{7}
Taking reciprocal of both the sides, we have
v100=76\dfrac{v}{{100}} = \dfrac{7}{6}
Multiplying both the sides by 100100, we finally get
v=7006v = \dfrac{{700}}{6}
v=166.67km/h\Rightarrow v = 166.67km/h

Hence, the required speed of the train to travel the next 100km100km is equal to 166.67km/h166.67km/h.

Note: Do not convert the distances and the speeds given in this question to their SI units, since all of them belong to the single system of units. In this question, the motion of the train was unidirectional. That’s why we calculated the average speed, otherwise we would have calculated the average velocity for which we would have considered the total displacement of the train.