A train covered the distance of 400 km between A and B at a certain speed. On the way back it covered 2 5 of the distance at that same speed and then it decreased its speed by 20 km/hour. Find the speed of the train at the end of its journey from B back to A, if the entire trip took 11 hours.

Initial velocity of train, V2 = 60m/s

Explanation:correct parameter in question is on the way back, the train covered 2/5 of the distance.

The distance travelked by an object with constant velocity is given by:

d = V * t

Distance, d = 400km

Decreased speed, V2 = V1 - 20km/h

V1 = initial velocity

On the way back, the train covered 2/5 of the entire distance with V1 and switched to V2.

Total time taken for the entire trip = 11hours

Let t1 and t2 be the times it took for the trip.

t1 = 400/V1

t2 = (2/5 * 400) / V1 + (3/5*400) / V2

t2 takes this form because the first 2/5 of the distance, the train travelled with V1 speed and the rest with V2

Therefore, t1 + t2 = to

400/V1 + (2/5*400) / V1 + (3/5*400) / V2

Multiplying each element by V (V1 - 20)

Leaves us with:

11 (V1-20) V1 = 400 (V1-20) + 160 (V1-20) + 240V1

11V1^2 - 1020V1 + 11200 = 0

Using quadratic formula to solve for V1

[ - b + - Sqrt (b^2 - 4ac) ] / 2a

[1020 + - Sqrt (1020^2 - 4 * 11 * 11200) ] / (2*11)

V1 = 80kmh or 114/11 kmh

On the way back, the train's speed changed by 20kmh 140/11>20kmh

This cannot be the initial speed of the train because if the speed dropped by 20kmh, the speed becomes negative.

Therefore, V1 = 80kmh

V2 = V1 - 20

V2 = 80 - 20

V2 = 60kmh