A subway train starts from rest at a station and acceleratesat arateof1.60 m/s
2
for 14.0 s. It runs at a constant speed for 70.0 s andslowsdownat a rate of 3.50 m/s
2 until it stops at the next station. Findthetotaldistance covered.

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

1. **Acceleration phase**: The train accelerates from rest.
2. **Constant speed phase**: The train moves at constant speed.
3. **Deceleration phase**: The train slows down until it stops.

We'll calculate the distance covered in each phase and then sum them up.

---

### 1. **Acceleration Phase**
- Initial velocity \( u_1 = 0 \, \text{m/s} \) (since the train starts from rest)
- Acceleration \( a_1 = 1.60 \, \text{m/s}^2 \)
- Time of acceleration \( t_1 = 14.0 \, \text{s} \)

The velocity at the end of this phase (which will be the constant speed for the next phase) can be calculated using:
\[
v = u_1 + a_1 t_1
\]
\[
v = 0 + (1.60)(14.0) = 22.4 \, \text{m/s}
\]

Now, the distance covered during the acceleration phase is:
\[
d_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2
\]
\[
d_1 = 0 + \frac{1}{2} (1.60)(14.0)^2 = 156.8 \, \text{m}
\]

---

### 2. **Constant Speed Phase**
- The velocity during this phase is the same as the final velocity of the acceleration phase, \( v = 22.4 \, \text{m/s} \)
- Time at constant speed \( t_2 = 70.0 \, \text{s} \)

The distance covered at constant speed is:
\[
d_2 = v t_2 = (22.4)(70.0) = 1568 \, \text{m}
\]

---

### 3. **Deceleration Phase**
- Initial velocity \( u_3 = 22.4 \, \text{m/s} \) (the same as the final velocity of the previous phase)
- Final velocity \( v_3 = 0 \, \text{m/s} \) (since the train stops)
- Deceleration \( a_3 = 3.50 \, \text{m/s}^2 \)

We can calculate the time taken to stop during deceleration using:
\[
v_3 = u_3 - a_3 t_3
\]
\[
0 = 22.4 - (3.50) t_3
\]
\[
t_3 = \frac{22.4}{3.50} = 6.4 \, \text{s}
\]

The distance covered during deceleration is:
\[
d_3 = u_3 t_3 - \frac{1}{2} a_3 t_3^2
\]
\[
d_3 = (22.4)(6.4) - \frac{1}{2} (3.50)(6.4)^2
\]
\[
d_3 = 143.36 - 71.68 = 71.68 \, \text{m}
\]

---

### Total Distance Covered
The total distance is the sum of the distances from all three phases:
\[
d_{\text{total}} = d_1 + d_2 + d_3
\]
\[
d_{\text{total}} = 156.8 + 1568 + 71.68 = 1796.48 \, \text{m}
\]

Thus, the total distance covered by the subway train is **1796.48 meters**.

---

Do you want more details, or have any questions?

Here are 5 related questions:
1. What is the maximum velocity the train reaches during the trip?
2. How much time does the entire journey take?
3. What would be the effect on the total distance if the acceleration rate changed?
4. How would the deceleration phase change if the final velocity wasn't zero?
5. How can we apply similar calculations to projectile motion?

**Tip**: Always break down a problem with multiple phases into smaller, manageable parts to simplify calculations.

A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s² for 14.0 s. It runs at a constant speed for 70.0 s and slows down at a rate of 3.50 m/s² until it stops at the next station. Find the total distance covered.

Learn how to find the total distance covered by a subway train starting from rest, accelerating at 1.60 m/s² for 14 seconds, running at constant speed for 70 seconds, and decelerating at 3.50 m/s² until it stops. This solution breaks down the problem using kinematic equations of motion, suitable for students in Grades 9-12.

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