Finding the total distanced covered (physics)

Question

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

Find the total distance covered. State your answer in units of km.

We are learning UAM equations in physics and I just don't know how to begin with this. The question leads me to believe I need to integrate, but I'm not too sure.

Any tips as to how I can solve this are greatly appreciated!

Answer 1

I will assume that these are the equations you need to use to solve this problem.

To find the total distance covered is to find the sum of the distances it travels

1. during acceleration
2. at constant speed
3. during deceleration

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Only three pieces of information about how the train accelerates are given:

1. initial speed (0 m/s)
2. acceleration ($1.60\frac{m}{s^2}$)
3. time ($14.0s$)

Using two equations above you will be able find out the distance it travels during this period of acceleration and its final speed, which can be used to calculate the distance it travels at constant speed (with the $70.0s$).

And then what you've left is just the distance traveled during deceleration.

Answer 2

The first phase accelerates for some time so you can compute the distance travelled and the speed at the end of this phase.

The speed is constant for the second phase and you know the duration.

For the final phase, you know the speed and deceleration so you can compute the time until the train stops. This allows you to compute the distance taken to stop.

Add the three distances together.

Answer 3

By integrating, you can show that the speed of the train is $v=1.6t$ for $0\leq t\leq 14$, $v=1.6(14)$ for $14\leq t\leq 84$ and $v=1.6(14)-3.5t$ for $84\leq t\leq \dfrac{1.6(14)}{3.5}$. Integrating again will give you the distance covered.