A 85-m
-long train begins uniform acceleration from rest. The front of the train has a speed of 36 m/s
 when it passes a railway worker who is standing 180 m
 from where the front of the train started. What will be the speed of the last car as it passes the worker? (See  (Figure 1))
Express your answer with the appropriate units.

To solve this problem, we need to calculate the speed of the last car of the train as it passes the railway worker, given that the train has been uniformly accelerating from rest.

### Step-by-Step Solution:

1. **Given Data:**
   - Length of the train, \( L = 85 \, \text{m} \)
   - Speed of the front of the train when it passes the worker, \( v_{\text{front}} = 36 \, \text{m/s} \)
   - Distance of the railway worker from the starting point of the front of the train, \( d = 180 \, \text{m} \)
   - Initial speed of the train, \( u = 0 \, \text{m/s} \) (since it starts from rest)

2. **Determine the acceleration:**
   Using the kinematic equation that relates velocity, acceleration, and distance:
   \[
   v^2 = u^2 + 2ad
   \]
   where:
   - \( v = v_{\text{front}} = 36 \, \text{m/s} \)
   - \( u = 0 \, \text{m/s} \)
   - \( d = 180 \, \text{m} \)

   Solving for acceleration \( a \):
   \[
   (36)^2 = 0 + 2a(180)
   \]
   \[
   1296 = 360a
   \]
   \[
   a = \frac{1296}{360} = 3.6 \, \text{m/s}^2
   \]

3. **Calculate the distance covered by the last car when it reaches the worker:**
   The last car is \( 85 \, \text{m} \) behind the front of the train. So, when the front of the train passes the worker, the last car is at a distance of:
   \[
   180 \, \text{m} - 85 \, \text{m} = 95 \, \text{m}
   \]
   This means the last car will have to travel \( 180 + 85 = 265 \, \text{m} \) in total to pass the worker.

4. **Determine the speed of the last car when it passes the worker:**
   We use the same kinematic equation, but now the distance is \( 265 \, \text{m} \):
   \[
   v^2 = u^2 + 2ad
   \]
   where:
   - \( u = 0 \, \text{m/s} \) (the train starts from rest)
   - \( a = 3.6 \, \text{m/s}^2 \)
   - \( d = 265 \, \text{m} \)

   Substituting these values:
   \[
   v^2 = 0 + 2(3.6)(265)
   \]
   \[
   v^2 = 1908
   \]
   \[
   v = \sqrt{1908} = 43.68 \, \text{m/s}
   \]

Thus, the speed of the last car as it passes the worker is approximately \( 43.68 \, \text{m/s} \).

### Final Answer:
The speed of the last car as it passes the worker is \( \boxed{43.7 \, \text{m/s}} \).

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Would you like further details or clarification on any part of the solution?

Here are 5 related questions to think about:
1. How would the result change if the train length were doubled?
2. What if the train had a non-uniform acceleration?
3. Can you derive the time it took for the train to reach the worker?
4. How would the stopping distance change if the train were to decelerate uniformly?
5. How does the mass of the train affect the acceleration, assuming constant force?

**Tip:** Always make sure to check the units and physical meaning of your answer after using kinematic equations!

A 85-m-long train begins uniform acceleration from rest. The front of the train has a speed of 36 m/s when it passes a railway worker who is standing 180 m from where the front of the train started. What will be the speed of the last car as it passes the worker? (See Figure 1) Express your answer with the appropriate units.

Solve a physics problem involving a 85-meter-long train that accelerates uniformly from rest. Learn how to calculate the speed of the last car as it passes a railway worker. The train's front reaches a speed of 36 m/s after traveling 180 meters. Suitable for high school students in Grades 11-12.

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