The problem in the image is about two trains moving towards each other on the same track with different speeds. Here’s a summary of the information:

1. The speed of the first train is 72 km/h.
2. The speed of the second train is 80 km/h.
3. When the train drivers notice each other, the distance between the two trains is 1 km.
4. The first train reduces its speed with a deceleration of 5184 km/h².
5. The question asks: What is the minimum deceleration that the second train must apply to avoid a collision?

Solution:

We will solve this using kinematic equations. Let’s define the following variables:

• $v_1$ = 72 km/h (speed of the first train)
• $v_2$ = 80 km/h (speed of the second train)
• $a_1$ = -5184 km/h² (deceleration of the first train, negative because it’s slowing down)
• $a_2$ = ? (deceleration of the second train, which we need to find)
• $d$ = 1 km (distance between the two trains)

Step 1: Convert speeds to consistent units (km/h to m/s)

To work in SI units, we need to convert the speeds from km/h to m/s:

• $v_1 = 72 \, \text{km/h} = 72 \times \frac{1000}{3600} = 20 \, \text{m/s}$
• $v_2 = 80 \, \text{km/h} = 80 \times \frac{1000}{3600} = 22.22 \, \text{m/s}$

Step 2: Convert deceleration to m/s²

We also need to convert the deceleration of the first train to m/s²:

• $a_1 = 5184 \, \text{km/h}^2 = 5184 \times \left( \frac{1000}{3600} \right)^2 \approx 400 \, \text{m/s}^2$

Step 3: Apply the relative motion and kinematic equation

The two trains are moving towards each other, so their relative velocity is: $v_{\text{relative}} = v_1 + v_2 = 20 + 22.22 = 42.22 \, \text{m/s}$

The combined distance they need to cover is 1 km or 1000 meters. For a collision to be avoided, the stopping distance of both trains combined should equal this distance.

Using the kinematic equation for stopping distance $d = \frac{v^2}{2a}$, we can set up the equation for both trains:

For the first train: $d_1 = \frac{v_1^2}{2a_1} = \frac{20^2}{2 \times 400} = \frac{400}{800} = 0.5 \, \text{m}$

For the second train, the stopping distance is: $d_2 = \frac{v_2^2}{2a_2}$

Since the total distance between the trains is 1000 m: $d_1 + d_2 = 1000$

Substitute $d_1 = 0.5$: $0.5 + \frac{v_2^2}{2a_2} = 1000$ $\frac{22.22^2}{2a_2} = 999.5$ $a_2 = \frac{22.22^2}{2 \times 999.5} \approx 0.247 \, \text{m/s}^2$

Step 4: Convert back to km/h²

Now, convert the deceleration $a_2$ back to km/h²: $a_2 = 0.247 \, \text{m/s}^2 \times \left( \frac{3600}{1000} \right)^2 = 3195.84 \, \text{km/h}^2$

Final Answer:

The second train needs to decelerate at least $3195.84 \, \text{km/h}^2$ to avoid a collision.

Let me know if you need further clarification, or if you have more questions.

Related Questions:

1. How do you calculate stopping distance for a vehicle using kinematic equations?
2. What is the importance of relative velocity in collision avoidance problems?
3. Why do we need to convert units when solving real-world physics problems?
4. How does deceleration impact stopping time and distance?
5. What are other real-world scenarios where train deceleration calculations are important?

Tip: Always make sure units are consistent when performing physics calculations.