# What is the average speed of a train that travells between two places. 3/4 of distance is covered at speed of 80 km / h and 1/4 at 160 km / h

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I assumed the trip was 100 km. I determined 75 miles of the trip was traveled at a rate of 80 km/h. Using d = rt I determined that 75/80 took .9375 of an hour or 56.25 minutes.

I then determined the time it took to go the other 1/4 distance (or in my case 25 miles) which was 25/160 or .15625 hours or 9.375 minutes.

I now have the total time of my trip in minutes or hours.

The total distance I assumed to be 100 km, the total time = 65.625 minutes , therefore 100/65.625 is approximately - to 1.524 km per minute or approximately 91.43 km per hour.

We need to find the average speed of a train that travels between two points. Let the distance between the points be P.

Now speed = distance / time or time = distance/ speed

As 3/4 of the distance is covered at 80 km/ h

=> time taken (T1) = (3P/4) / 80 = 3P/ 320

The rest of the 1/4 of the distance is covered at 160 km/h

=> time taken (T2) = (P/4) / 160 = P/ 640

Therefore the total time taken is T1+ T2 = 3P/320 + P/ 640 = 7P / 640.

Therefore the average speed is total distance/ total time

=> P / ( 7P / 640 )

=> 640 /7

=> 91.42 km/h

**The required average speed is 91.42 km/h**

The train travels between 2 places, covering the distance that we'll note as x.

We'll split the x distance in 3/4, since the train covers the first 3/4 of x at the v1 speed, of 80 km/h, and the other 1/4 at the v2 speed, of 160 km/h.

The 3/4 of x distance is covered in the time t1, and the other 1/4, in the time t2.

We'll write the formula of speed:

v = x/t (1)

v1 = (3x/4)/t1

We'll substitute v1:

80 = (3x/4)/t1

320t1 = 3x

v2 = (x/4)/t2

We'll substitute v2:

160 = (x/4)/t2

t2 = x/640 hour

Now, we'll write the average speed:

av. v = total distance covered/total time taken

av. v = x/(t1+t2)

We'll substitute t1 and t2:

av. v = x/(3x/320+x/640)

av. v = 640/7

av. v = 92 Km/h

**The average speed of the train is av. v = 92 Km/h.**