Answer:
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