A subway train starts from rest at a station and accelerates at a rate of 1.50m/s^2 for 13.3s . It runs at constant speed for 71.0s and slows down at a rate of 3.48m/s^2 until it stops at the next station.

Find the total distance covered.
Express your final answer in kilometers

d1 = Vo*t + 0.5a*t^2,

di = 0 + 0.75*(13.3)^2 = 132.7 m.
Vf = at = 1.5m/s^2 * 13.3s = 19.95 m/s.

Vo = 19.95 m/s.
Vf^2 = Vo^2 + 2a*d,
d2 = (Vf^2-Vo^2) / 2a,
d2 = (0-(19.95)^2 /-3.48 = 114.4 m.

D = d1 + d2 = 132.7 + 114.4 = 247 m. =
0.247 km. = Tot. dist.

Correction: d2 = (0-(19.95)^2 / -6.96 =

57.2 m.

D = di + d2 = 132.7 + 57.2 = 190 m. =
0.19 km. = Tot. dist.

Since the train ran at constant speed for 71 seconds, we must add d3:

d3=Vo*t = 19.95m/s * 71s. = 1416.45 m.

D = d1 + d2 + d3,
D = 132.7 + 57.2 + 1416.45 = 1606.4 m. = 1.61 km = Tot. dist.

To find the total distance covered by the subway train, we need to analyze each phase of its motion and calculate the distance traveled during that phase.

Phase 1: Acceleration

The subway train starts from rest and accelerates at a rate of 1.50 m/s^2 for 13.3 seconds. We can use the kinematic equation:

distance = (initial velocity × time) + (0.5 × acceleration × time^2)

In this case, the initial velocity (u) is 0 m/s since the train starts from rest, the time (t) is 13.3 seconds, and the acceleration (a) is 1.50 m/s^2.

distance1 = (0 × 13.3) + (0.5 × 1.50 × 13.3^2)

Phase 2: Constant Speed

The subway train runs at a constant speed for 71.0 seconds. Since the speed is constant, the distance covered during this phase is given by:

distance2 = speed × time

However, we need to convert the time from seconds to kilometers per hour (km/h) to match the final answer units. 1 kilometer is equal to 1000 meters, and 1 hour is equal to 3600 seconds.

distance2 = speed × (71.0/3600) (km/h)

Phase 3: Deceleration

The subway train slows down at a rate of 3.48 m/s^2 until it stops at the next station. Again, we can use the kinematic equation to find the distance covered:

distance = (initial velocity × time) + (0.5 × acceleration × time^2)

In this case, the initial velocity (u) is the final velocity from phase 2 since the train continues at the same deceleration rate. We can calculate this final velocity:

final velocity = initial velocity + (acceleration × time)

The time (t) is the deceleration time from phase 3, which is not mentioned in the question, so we'll represent it as t3.

distance3 = (final velocity × t3) + (0.5 × deceleration × t3^2)

Finally, we can calculate the total distance covered by the subway train:

total distance = distance1 + distance2 + distance3

Now let's substitute the given values into the equations and calculate the distance.

distance1 = (0 × 13.3) + (0.5 × 1.50 × 13.3^2)

distance2 = speed × (71.0/3600) (km/h)

distance3 = (final velocity × t3) + (0.5 × deceleration × t3^2)

total distance = distance1 + distance2 + distance3

Please provide the values for the speed of the constant speed phase, as well as the final velocity and time for the deceleration phase so that we can compute the total distance covered.