A subway train starts from rest at a station and accelerates at a rate of
1.60 m/s2 for 8s. It runs at a constant speed for 70 s and decelerates at
2.5 m/s2 until it stops at the next station. Find the total distance (traveled)
covered.

To find the total distance covered by the subway train, we can break the motion into three parts:

1. **Acceleration phase**: The train accelerates from rest at \( 1.60 \, \text{m/s}^2 \) for 8 seconds.
2. **Constant speed phase**: The train runs at a constant speed for 70 seconds.
3. **Deceleration phase**: The train decelerates at \( 2.5 \, \text{m/s}^2 \) until it comes to rest.

### Step 1: Distance during acceleration phase

The formula to calculate the distance during acceleration is:
\[
d_1 = v_0 t + \frac{1}{2} a t^2
\]
where:
- \( v_0 = 0 \, \text{m/s} \) (initial velocity since the train starts from rest),
- \( a = 1.60 \, \text{m/s}^2 \),
- \( t = 8 \, \text{seconds} \).

Substitute the values:
\[
d_1 = 0 \cdot 8 + \frac{1}{2} (1.60) (8)^2 = 0 + 0.80 \cdot 64 = 51.2 \, \text{m}.
\]

At the end of the acceleration phase, the train's velocity \( v_f \) can be calculated using the formula:
\[
v_f = v_0 + a t = 0 + (1.60)(8) = 12.8 \, \text{m/s}.
\]

### Step 2: Distance during constant speed phase

During this phase, the train moves at a constant speed \( v = 12.8 \, \text{m/s} \) for 70 seconds. The distance traveled during this phase is:
\[
d_2 = v t = (12.8) (70) = 896 \, \text{m}.
\]

### Step 3: Distance during deceleration phase

The train decelerates at \( a = -2.5 \, \text{m/s}^2 \) until it comes to rest. To find the time taken to stop, use the formula:
\[
v_f = v_0 + a t.
\]
Setting \( v_f = 0 \) and \( v_0 = 12.8 \, \text{m/s} \), we solve for \( t \):
\[
0 = 12.8 + (-2.5) t \implies t = \frac{12.8}{2.5} = 5.12 \, \text{seconds}.
\]

The distance covered during this deceleration phase is:
\[
d_3 = v_0 t + \frac{1}{2} a t^2 = (12.8)(5.12) + \frac{1}{2} (-2.5) (5.12)^2.
\]
First, calculate the terms:
\[
d_3 = 65.536 + \frac{1}{2} (-2.5) (26.2144) = 65.536 - 32.768 = 32.768 \, \text{m}.
\]

### Total distance traveled

Now, sum the distances from all three phases:
\[
d_{\text{total}} = d_1 + d_2 + d_3 = 51.2 + 896 + 32.768 = 979.968 \, \text{m}.
\]

Thus, the total distance covered by the subway train is approximately **980 meters**.

---

Would you like further details or have any questions about the solution?

Here are 5 related questions:
1. How would the distance change if the acceleration were increased to \( 2 \, \text{m/s}^2 \)?
2. How does the deceleration rate affect the stopping distance?
3. Can you calculate the total time taken for the entire journey?
4. What if the constant speed phase lasted for 100 seconds instead of 70 seconds?
5. How would the problem change if the initial velocity was not zero?

**Tip**: In problems involving multiple phases of motion, breaking them into distinct sections and solving them step by step makes the solution clearer.

A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s2 for 8s. It runs at a constant speed for 70 s and decelerates at 2.5 m/s2 until it stops at the next station. Find the total distance (traveled) covered.

Solve a kinematics problem involving a subway train's motion with acceleration, constant velocity, and deceleration. This detailed solution covers key concepts in kinematics using the equations of motion, ideal for students in Grades 9-12. The total distance traveled is approximately 980 meters.

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