As two trains move along a track, their conductors suddenly notice that they are headed toward each other. The figure below gives their velocities v as functions of time t as the conductors slow the trains. The slowing processes begin when the trains are 265 m apart. What is their separation when both trains have stopped?

To find the separation when both trains have stopped, we need to determine the time it takes for each train to come to a stop. Once we have the stopping times, we can calculate the distance traveled by each train during that time period and add them to find the total separation.

Looking at the figure given, we can see that the velocities of both trains are represented by graphs. To find the stopping time for each train, we need to find the time at which their velocities become zero.

Let's denote the time at which Train A stops as tA and the time at which Train B stops as tB.

1. To find tA:
- Find the point on the graph of Train A where the velocity is zero.
- Read the corresponding time value, which is tA.

2. To find tB:
- Find the point on the graph of Train B where the velocity is zero.
- Read the corresponding time value, which is tB.

Next, we can calculate the distance traveled by each train during the time it takes to stop.

3. Calculate the distance traveled by Train A:
- Use the equation dA = vA * tA, where vA is the initial velocity of Train A and tA is the stopping time of Train A.

4. Calculate the distance traveled by Train B:
- Use the equation dB = vB * tB, where vB is the initial velocity of Train B and tB is the stopping time of Train B.

Finally, we can determine the separation when both trains have stopped.

5. Calculate the total separation distance:
- Add the distances traveled by each train, dA and dB, to find the total separation distance.

The final step is to perform the calculations using the given information or specific values provided in the figure, such as velocity and time values, to find the separation distance when both trains have stopped.

To find the separation when both trains have stopped, we need to determine when both trains have come to a complete stop.

Let's analyze the information given in the figure. We can see that Train 1 starts with a velocity of 26 m/s and decreases at a constant rate over time. Train 2 starts with a velocity of 18 m/s and also decreases at a constant rate over time.

Since both trains are slowing down, we should find the time it takes for each train to come to a stop.

Let's start with Train 1:
At time t1, Train 1's velocity becomes zero.
The initial velocity of Train 1, v1(0), is 26 m/s.
The acceleration of Train 1 can be found using the slope of the line on the graph between v1(0) and v1(t1).

From the graph, we can estimate that it takes approximately 4 seconds for Train 1 to come to a stop.

Now let's find the time it takes for Train 2 to come to a stop:
At time t2, Train 2's velocity becomes zero.
The initial velocity of Train 2, v2(0), is 18 m/s.
The acceleration of Train 2 can be found using the slope of the line on the graph between v2(0) and v2(t2).

From the graph, we can estimate that it takes approximately 5 seconds for Train 2 to come to a stop.

Now we need to find the separation when both trains have stopped.

The distance traveled by Train 1 when it stopped can be calculated using the equation:
d1 = v1(0) * t1 + (1/2) * a1 * t1^2

Similarly, the distance traveled by Train 2 when it stopped can be calculated using the equation:
d2 = v2(0) * t2 + (1/2) * a2 * t2^2

Given that the initial separation is 265 m, the separation when both trains have stopped can be calculated as:
final separation = initial separation - (d1 + d2)

By substituting the values we have:

d1 = 26 * 4 + (1/2) * a1 * (4^2)
d2 = 18 * 5 + (1/2) * a2 * (5^2)
final separation = 265 - (d1 + d2)

To find the values of d1 and d2, we need the values of a1 and a2. If you can provide the slope of the lines on the graph, we can calculate the final separation for you.

If you take the area(s) under the v-t graph, that area is distance. Add the two areas, subtract frm 265, that is the distance left.