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.

We can break the motion of the subway train into three stages:

1. **Acceleration phase:** The train accelerates at \( a_1 = 1.60 \, \text{m/s}^2 \) for \( t_1 = 14.0 \, \text{s} \).
2. **Constant speed phase:** The train moves at a constant speed for \( t_2 = 70.0 \, \text{s} \).
3. **Deceleration phase:** The train decelerates at \( a_3 = 3.50 \, \text{m/s}^2 \) until it comes to rest.

### Step 1: Distance during acceleration

The initial velocity \( u_1 = 0 \) (since the train starts from rest), and the acceleration is \( a_1 = 1.60 \, \text{m/s}^2 \). The velocity at the end of this phase is:

\[
v_1 = u_1 + a_1 t_1 = 0 + (1.60 \, \text{m/s}^2)(14.0 \, \text{s}) = 22.4 \, \text{m/s}
\]

The distance covered during this phase can be found using the equation:

\[
d_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2 = 0 + \frac{1}{2} (1.60 \, \text{m/s}^2) (14.0 \, \text{s})^2
\]

\[
d_1 = \frac{1}{2} (1.60)(196) = 156.8 \, \text{m}
\]

### Step 2: Distance during constant speed

The velocity during this phase is the same as \( v_1 = 22.4 \, \text{m/s} \). The distance covered during this phase is:

\[
d_2 = v_1 t_2 = (22.4 \, \text{m/s}) (70.0 \, \text{s}) = 1568 \, \text{m}
\]

### Step 3: Distance during deceleration

In this phase, the train decelerates to rest. The final velocity \( v_3 = 0 \), the initial velocity \( v_2 = 22.4 \, \text{m/s} \), and the deceleration is \( a_3 = -3.50 \, \text{m/s}^2 \). Using the equation:

\[
v_3^2 = v_2^2 + 2 a_3 d_3
\]

Solving for \( d_3 \):

\[
0 = (22.4 \, \text{m/s})^2 + 2 (-3.50 \, \text{m/s}^2) d_3
\]

\[
d_3 = \frac{(22.4)^2}{2(3.50)} = \frac{501.76}{7.0} = 71.68 \, \text{m}
\]

### Total distance covered

Now, we sum up the distances from all three phases:

\[
d_{\text{total}} = d_1 + d_2 + d_3 = 156.8 \, \text{m} + 1568 \, \text{m} + 71.68 \, \text{m} = 1796.48 \, \text{m}
\]

Thus, the total distance covered by the train is approximately **1796.5 meters**.

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Would you like more details or have any questions?

Here are 5 related questions for further practice:
1. How would the total distance change if the acceleration rate was increased?
2. What is the total time taken for the journey?
3. How much time would it take for the train to decelerate if the deceleration was reduced to 2.0 m/s²?
4. What would be the distance covered if the train maintained a higher constant speed for the same duration?
5. How would the problem change if the initial speed was not zero?

**Tip:** Always break a motion problem into phases—acceleration, constant speed, and deceleration—to simplify the calculation.

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.

Calculate the total distance covered by a subway train as it accelerates at 1.60 m/s² for 14.0 s, moves at a constant speed for 70.0 s, and decelerates at 3.50 m/s² until stopping. This detailed kinematic solution is suitable for Grades 9-12, involving equations of motion and key concepts in physics.

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