A subway train starts from rest at a station and accelerates at a rate of 1.60m/s^2 for 14.0 s. It runs at constant speed for 70.0 s and slows down at a rate of 3.50m/s^2 until it stops at the next station.

What is the total distance covered?

distance:

first leg: avg velocity= 1/2 1.6*14
second leg: avg velocity = 1.6*14
third leg: avg veloctiy= 1/2 * (1.6*14)

multiply each of these by the time for the leg, then add the distances.

the time for the last leg can be found from Vf=Vi+at...

I am really confused as to what you are doing here.

To find the total distance covered, we need to calculate the distance covered during each phase of the train's motion and then add them together.

1. Distance covered in the acceleration phase:
To calculate the distance covered during the acceleration phase, we can use the equation:
distance = initial velocity × time + 0.5 × acceleration × time^2
Since the train starts from rest, the initial velocity is 0 m/s. The acceleration is given as 1.60 m/s^2, and the time is given as 14.0 s. Plugging these values into the equation, we get:
distance = 0 × 14.0 + 0.5 × 1.60 × (14.0)^2

2. Distance covered at constant speed:
During this phase, the train runs at a constant speed, so the distance covered is given by the formula:
distance = speed × time
The time for this phase is given as 70.0 s, and since the speed remains constant, we need to find the speed of the train during this phase.

3. Distance covered during deceleration:
Similar to the acceleration phase, we can use the same distance formula, but this time with a negative acceleration since the train is slowing down. The acceleration is given as -3.50 m/s^2 and the time is the same as the acceleration phase (14.0 s).

Now, to calculate the total distance covered, we need to add up the distances from each phase:
Total distance = distance in acceleration phase + distance at constant speed + distance in deceleration phase

Let's calculate each part to find the total distance covered.