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

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