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NoPhysicsGenius
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I am having difficulties solving the first part of Problem 83 from Chapter 2 of Physics for Scientists and Engineers by Paul A. Tipler, 4th edition. The statement of the problem is as follows:
My attempted solution is as follows:
The two trains will continue to travel at their respective constant velocities for 0.4 s. (Thereafter, the engineer of the passenger train will apply the brakes.)
Displacement(xp) = vp * t = (29 m/s) * 0.4 s = 11.6 m
Displacement(xf) = vf * t = (6 m/s) * 0.4 s = 2.4 m
After the first 0.4 s, the distance between the passenger train and the freight train becomes 360 m - (11.6 m - 2.4 m) = 350.8 m
Now the engineer of the passenger train applies the brakes ...
The freight train has the following equation of motion:
xf - xf0 = vf0 * t
xf0 = 350.8 m
vf0 = 6 m/s
Therefore, xf = (vf0 * t) + xf0 = (6 * t) + 350.8
The passenger train has the following equation of motion:
xp - xp0 = (vp0 * t) + (0.5 * ap * t2)
xp0 = 0
vp0 = 29 m/s
Therefore, xp = (29 * t) + (0.5 * ap * t2)
Now vp = vp0 + (ap * t) => ap = (vp - vp0) / t
vp = 0 (The passenger train will eventually come to a rest.)
vp0 = 29 m/s
Therefore, ap = (0 - 29) / t = -29 / t
Because we want to avoid a collision, it must be that xp < xf.
Therefore, (29 * t) + (0.5 * ap * t2) < (6 *t) + 350.8
=> (0.5 * ap * t2) + (23 * t) - 350.8 < 0
Substituting ap = -29 / t yields:
[0.5 * (-29 / t) * t2] + (23 * t) - 350.8 < 0
=> [(-29 / 2) * t] + (23 * t) - 350.8 < 0
=> {[23 - (29 / 2)] * t} - 350.8 < 0
=> (8.5 * t) - 350.8 < 0
=> 8.5 * t < 350.8
=> t < 350.8 / 8.5
=> t < 41.3 s
Noting that ap = - 29 / t => t = - 29 / ap, we then have:
(-29 / ap) < 41.3
=> ap > (-29 / 41.3)
=> ap > -0.702 m/s2
There are two problems with this answer:
So, can anyone please spot the error(s) in my reasoning?
This is my first post here, so please be kind. Thank you.
A passenger train is traveling at 29 m/s when the engineer sees a freight train 360 m ahead traveling on the same track in the same direction. The freight train is moving at a speed of 6 m/s. If the reaction time of the engineer is 0.4 s, what must be the deceleration of the passenger train if a collision is to be avoided?
My attempted solution is as follows:
The two trains will continue to travel at their respective constant velocities for 0.4 s. (Thereafter, the engineer of the passenger train will apply the brakes.)
Displacement(xp) = vp * t = (29 m/s) * 0.4 s = 11.6 m
Displacement(xf) = vf * t = (6 m/s) * 0.4 s = 2.4 m
After the first 0.4 s, the distance between the passenger train and the freight train becomes 360 m - (11.6 m - 2.4 m) = 350.8 m
Now the engineer of the passenger train applies the brakes ...
The freight train has the following equation of motion:
xf - xf0 = vf0 * t
xf0 = 350.8 m
vf0 = 6 m/s
Therefore, xf = (vf0 * t) + xf0 = (6 * t) + 350.8
The passenger train has the following equation of motion:
xp - xp0 = (vp0 * t) + (0.5 * ap * t2)
xp0 = 0
vp0 = 29 m/s
Therefore, xp = (29 * t) + (0.5 * ap * t2)
Now vp = vp0 + (ap * t) => ap = (vp - vp0) / t
vp = 0 (The passenger train will eventually come to a rest.)
vp0 = 29 m/s
Therefore, ap = (0 - 29) / t = -29 / t
Because we want to avoid a collision, it must be that xp < xf.
Therefore, (29 * t) + (0.5 * ap * t2) < (6 *t) + 350.8
=> (0.5 * ap * t2) + (23 * t) - 350.8 < 0
Substituting ap = -29 / t yields:
[0.5 * (-29 / t) * t2] + (23 * t) - 350.8 < 0
=> [(-29 / 2) * t] + (23 * t) - 350.8 < 0
=> {[23 - (29 / 2)] * t} - 350.8 < 0
=> (8.5 * t) - 350.8 < 0
=> 8.5 * t < 350.8
=> t < 350.8 / 8.5
=> t < 41.3 s
Noting that ap = - 29 / t => t = - 29 / ap, we then have:
(-29 / ap) < 41.3
=> ap > (-29 / 41.3)
=> ap > -0.702 m/s2
There are two problems with this answer:
1. The magnitude for the acceleration of the passenger train is given in the back of the book as 0.754 m/s2, not 0.702 m/s2; and
2. The acceleration of the passenger train should be less than a negative quantity, not greater than a negative quantity. For example, if ap > -0.702 m/s2, then ap might be zero, in which case the passenger train very quickly catches up to the freight train (at a constant velocity) and collides with it.
2. The acceleration of the passenger train should be less than a negative quantity, not greater than a negative quantity. For example, if ap > -0.702 m/s2, then ap might be zero, in which case the passenger train very quickly catches up to the freight train (at a constant velocity) and collides with it.
So, can anyone please spot the error(s) in my reasoning?
This is my first post here, so please be kind. Thank you.