the engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200m ahead on the same track. The freight train is traveling at 15.0 m/s in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant accelaration of -0.100 m/s2, while the freight train continues with constant speed. Take x=0 at the location of the front of the passenger train when the engineer applies the brake. (a) will the cows nearby witness a collision? (b) if so, where it take place? (c) on a single graph, sketch the positions of the frong of the passenger train and the back of the freight train.

With no braking, the trains would collide in t = 200/(25-15) = 20 sec.

At a decelerauon of 0.10m/sec.^2, the passenger train would come to a stop in Vf = 0 = 25 - 0.10t yielding t = 250 sec.

Since the time to impact is less than the time to the passenger train stopping, the two trains will collide when the freight train has traveled 300 m. and the passenger train has traveled 500m.

To answer this question, we need to determine whether the passenger train will collide with the freight train and if so, at what location. We also need to sketch the positions of the front of the passenger train and the back of the freight train on a single graph.

Let's break down the problem step by step:

(a) To determine whether a collision will occur, we need to find the time it takes for the passenger train to reach the freight train.

First, let's find the relative velocity between the passenger and freight trains. We subtract the velocity of the freight train from the velocity of the passenger train:

Relative velocity = Velocity of passenger train - Velocity of freight train
Relative velocity = 25.0 m/s - 15.0 m/s = 10.0 m/s (in the same direction as the passenger train)

Now, we can find the time it takes for the passenger train to catch up to the freight train. We can use the formula:

Relative velocity = (final position - initial position) / time

In this case, the final position is 200 m (since the caboose of the freight train is 200 m ahead of the passenger train) and the initial position is 0 m (since we take x=0 as the location of the front of the passenger train when the brakes are applied).

Using this formula, we can rearrange it to solve for time:

time = (final position - initial position) / relative velocity
time = (200 m - 0 m) / 10.0 m/s
time = 20 s

Therefore, it will take 20 seconds for the passenger train to catch up to the freight train.

(b) To find where the collision will take place, we can calculate the distance traveled by both the passenger train and the freight train during this time.

Let's calculate the distance traveled by the passenger train:

The passenger train is initially traveling at 25.0 m/s but decelerates at a constant acceleration of -0.100 m/s². We can use the kinematic equation:

v² = u² + 2as

where:
v = final velocity (0 m/s, as the passenger train will come to a stop)
u = initial velocity (25.0 m/s)
a = acceleration (-0.100 m/s²)
s = distance (to be determined)

Rearranging the formula to solve for distance:

s = (v² - u²) / (2a)
s = (0 m/s)² - (25.0 m/s)² / (2 * -0.100 m/s²)
s = 3125 m / (-0.200 m/s²)
s = -15625 m

The negative sign indicates that the distance is in the opposite direction to the passenger train's initial velocity.

Now, let's calculate the distance traveled by the freight train:

The freight train is traveling at a constant speed of 15.0 m/s, so its distance traveled can be found using:

distance = velocity * time
distance = 15.0 m/s * 20 s
distance = 300 m

Since the passenger train has traveled -15625 m (opposite direction) and the freight train has traveled 300 m during the same time, we can determine that a collision will occur when their distances are equal in magnitude:

|-15625 m| = 300 m

As the left side of the equation is much larger than the right side, we can see that a collision will indeed occur.

To find where the collision will take place, we need to add the distance traveled by the passenger train to its initial position (0 m):

collision location = initial position + distance traveled by passenger train
collision location = 0 m + (-15625 m)
collision location = -15625 m

Therefore, the collision will take place at a location 15625 meters behind the initial position of the front of the passenger train when the brakes are applied.

(c) We can sketch the positions of the front of the passenger train and the back of the freight train on a single graph.

Assuming time is on the x-axis and position is on the y-axis, we can plot the positions as follows:

- Passenger Train: The position of the front of the passenger train starts at y=0 and decreases linearly with time until it reaches y=-15625 m at 20 seconds (when the collision occurs).
- Freight Train: The position of the back of the freight train remains constant at y=200 m throughout.

The graph would show the passenger train's position starting at 0 and decreasing linearly until it intersects with the y=200 m line at t=20 s. That intersection point marks the location of the collision.

To solve this problem, we need to analyze the motion of both trains and determine if they will collide.

(a) Will the cows nearby witness a collision?
To determine if the two trains will collide, we need to find the time it takes for the passenger train to stop.

We can use the formula for acceleration:
v = u + at
where:
v = final velocity (0 m/s in this case, since the passenger train stops)
u = initial velocity of the passenger train (25.0 m/s)
a = acceleration (-0.100 m/s^2)
t = time

Rearranging the equation to solve for time (t):
t = (v - u) / a
t = (0 - 25.0) / (-0.100)
t = 250 seconds

Therefore, it will take 250 seconds for the passenger train to stop.

(b) If there is a collision, where will it take place?
Since the passenger train is traveling at a higher speed than the freight train and the passenger train stops, there will be a collision.

To find the position where the collision occurs, we need to find the distance traveled by both trains during the time it takes for the passenger train to stop.

Distance traveled by the passenger train:
s = ut + (1/2)at²
s = (25.0)(250) + (1/2)(-0.100)(250)²
s = 6250 - 3125
s = 3125 meters

Distance traveled by the freight train:
s = ut
s = (15.0)(250)
s = 3750 meters

Therefore, the collision will take place when the passenger train has traveled 3125 meters from its starting point, and the freight train has traveled 3750 meters from its starting point.

(c) On a single graph, sketch the positions of the front of the passenger train and the back of the freight train:
Let's assume the x-axis represents distance (in meters) and the y-axis represents time (in seconds).

|
| Freight
|-------------------------
|
|
|
| Passenger
-----------------------------------
0 3125 3750

The graph shows the position of the front of the passenger train increasing linearly from 0 meters to 3125 meters, while the back of the freight train remains constant at 3750 meters. The collision will occur at the point where the two lines intersect, which happens at 3125 meters on the x-axis.