The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m 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 acceleration of , while the freight train continues with constant speed. Take at the location of the front of the passenger train when the engineer applies the brakes.
From the given information, we can see that the relative speed of the passenger train with respect to the freight train is V_passenger = 25 m/s - 15 m/s = 10 m/s.
Let's find the time (t) it takes for the passenger train to come to a stop.
v = u + at
0 = 25 + (-0.4)t
t = 25/0.4
t = 62.5 seconds
Now, let's find the distance (d) traveled by the passenger train during this time (t).
d = ut + (1/2)at^2
d = 25(62.5) + (1/2)(-0.4)(62.5^2)
d = 1562.5 - 781.25
d = 781.25 meters
Now, let's find the distance (d_f) traveled by the freight train during this time (t).
d_f = 15 * 62.5
d_f = 937.5 meters
Since the freight train was initially 200 meters ahead, we can find the remaining distance between the two trains by subtracting the distance traveled by the passenger train from the total distance traveled by the freight train.
Remaining_distance = 937.5 - 781.25
Remaining_distance = 156.25 meters
Therefore, the remaining distance between the two trains after the passenger train stops is 156.25 meters.
To find the location of the front of the passenger train when the engineer applies the brakes, we can use the equations of motion.
Given:
Initial velocity of the passenger train, u1 = 25.0 m/s
Distance between the caboose of the freight train and the front of the passenger train, s = 200 m
Velocity of the passenger train after the brakes are applied, v1 = 0 m/s
Acceleration of the passenger train, a = ?
Let's find the acceleration of the passenger train.
We can use the equation of motion: v^2 = u^2 + 2as
Rearranging the equation to isolate acceleration, we have:
a = (v^2 - u^2) / (2s)
Since the velocity after the brakes are applied is zero, v1 = 0 m/s. Plugging in the values, we have:
a = (0^2 - 25.0^2) / (2 * -200)
Calculating this,
a = -625 / (-400) = 1.5625 m/s^2
The acceleration of the passenger train is 1.5625 m/s^2.
Now, we can find the position of the front of the passenger train when the brakes are applied.
We can use the equation of motion: s = ut + (1/2)at^2
Rearranging the equation to solve for position, we have:
s = ut + (1/2)at^2
Since the final velocity is zero, v1 = 0 m/s. Plugging in the values, we have:
s = 25.0(0) + (1/2)(1.5625)t^2
Simplifying this equation to isolate t^2, we have:
2s = 1.5625t^2
Rearranging the equation, we have:
t^2 = (2s) / 1.5625
t^2 = (2 * 200) / 1.5625 = 256
Taking the square root of both sides, we have:
t = √256 = 16
The time taken by the passenger train to reach a velocity of 0 m/s is 16 seconds.
Now, we can find the position of the front of the passenger train using the equation of motion:
s = ut + (1/2)at^2
Plugging in the values, we have:
s = 25.0(16) + (1/2)(1.5625)(16)^2
s = 400 + 1000
s = 1400 m
Therefore, the position of the front of the passenger train when the engineer applies the brakes is 1400 meters.
To find the location of the front of the passenger train when the engineer applies the brakes, we can use the equations of motion. The equation we will use is:
\[y = y_0 + v_0t + \frac{1}{2}at^2\]
Where:
y is the final position of the front of the passenger train
y0 is the initial position of the front of the passenger train (200 m)
v0 is the initial velocity of the passenger train (25.0 m/s)
a is the constant acceleration of the passenger train (unknown)
t is the time taken for the passenger train to stop
We need to find the value of a. To do that, we can use another equation of motion:
\[v = v_0 + at\]
Where:
v is the final velocity of the passenger train (0 m/s)
v0 is the initial velocity of the passenger train (25.0 m/s)
a is the constant acceleration of the passenger train (unknown)
t is the time taken for the passenger train to stop
Solving this equation for a gives us:
\[a = \frac{-v_0}{t}\]
Now, we can substitute the value of a into the first equation to find the value of t when the passenger train stops:
\[0 = v_0t + \frac{1}{2}\bigg(\frac{-v_0}{t}\bigg)t^2\]
Simplifying this equation gives us:
\[0 = v_0t - \frac{1}{2}v_0t\]
\[0 = \frac{1}{2}v_0t\]
Since t cannot be 0 (because it takes time for the passenger train to stop), we can cancel out the t term, which gives us:
\[0 = \frac{1}{2}v_0\]
Multiplying both sides of the equation by 2 gives us:
\[0 = v_0\]
Since v0 is the initial velocity of the passenger train (25.0 m/s), this means that the initial velocity is 0 when the passenger train stops.
Now, we can substitute the values into the first equation to find the final position of the front of the passenger train when it stops:
\[y = y_0 + v_0t + \frac{1}{2}at^2\]
\[y = 200 + 0t + \frac{1}{2}\bigg(\frac{-v_0}{t}\bigg)t^2\]
\[y = 200 + 0 + \frac{1}{2}\bigg(\frac{-v_0}{t}\bigg)t^2\]
\[y = 200 + 0\]
Therefore, the location of the front of the passenger train when the engineer applies the brakes is 200 meters from the caboose of the freight train.