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Need Help on Basic Kinematics Problem

  1. Aug 15, 2004 #1
    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:

    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.​

    So, can anyone please spot the error(s) in my reasoning?

    This is my first post here, so please be kind. :biggrin: Thank you.
     
  2. jcsd
  3. Aug 16, 2004 #2

    cepheid

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    The passenger train doesn't have to come to a full stop before reaching the freight train to avoid a collision, it merely has to slow to 6m/s before reaching the freight train. Compute the answer again with [tex] \inline v_{p} = 6m/s [/tex], and you should obtain the correct answer. I'll have to think about the second problem you're having. The signs are confusing me.
     
  4. Aug 16, 2004 #3

    cepheid

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    Actually...

    It's quite simple: [tex] \inline a_p [/tex] is intrinsically negative because the passenger train is decelerating. Since you multiplied both sides of the inequality by [tex] \inline a_p [/tex] in the above step, you have to flip the sign of the inequality.
     
  5. Aug 16, 2004 #4

    Clausius2

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    Maybe the time you have spent writing such ellaborate thread you could have re-written your solution of the problem, and surely you would find the error. :wink:
     
  6. Aug 16, 2004 #5
    Attempt to solve the problem using Relative Velocity

    NoPhysicsGenius, welcome to PF!

    You can do this problem another way:

    [tex]v_{p,i} = +29 m/s[/tex]
    [tex]v_{f,i} = +6 m/s[/tex]
    (both velocities with respect to a stationary ground observer; + sign implies direction of motion, - sign implies opposite to direction of motion of freight and passenger trains--sorry but I haven't figured out how to do caps on unit vectors with TeX here, yet)

    subscripts: i = initial, f = freight train, p = passenger train

    Reaction time means the time elapsed between first sighting the train and applying the brakes (imposing deceleration). If I use relative velocity to analyze the situation, I can say that the velocity of the passenger train relative to the freight train is given by

    [tex]v_{p/f,i} = v_{p,i} - v_{f,i} = (+29 m/s) - (+6 m/s) = +23 m/s [/tex]

    This 'trick' can allow me write just one equation of motion because I am considering the freight train to be fixed and the velocity of the passenger train with respect to an observer on the freight train is +23 m/s. It is with this velocity (of approach) that the passenger train covers the 360 metre distance. This distance is effectively shortened because the driver takes 0.4 seconds to figure out that he is going to have a bad day ;-), so the effective distance covered by the passenger train during the phase involving deceleration is simply

    s = 360m - (23 m/s)(0.4s) = (360 - 9.2)m = 350.8m

    This is the maximum distance that the train can cover to "just" (marginally) avoid a collision. If I denote the deceleration imposed by the brakes as a, then

    [tex]
    v^{2} = u^{2} - 2as
    [/tex]

    gives

    [tex]
    s <= \frac{v_{p/f,i}^{2}}{2a}
    [/tex]

    or

    [tex]
    a <= \frac{v_{p/f,i}^{2}}{2s}
    [/tex]

    Plug in the values and I think you should get the correct answer.

    Note: I am sorry if this post appears to be an exercise in spoonfeeding. I went ahead and posted this solution solely to demonstrate its slight algebraic sophistication (in time and effort mainly and not in the concept) over the conventional approach not involving relative velocity. Again, you shouldn't except it without debate and thought.

    Hope that helps,

    cheers
    vivek
     
    Last edited: Aug 16, 2004
  7. Aug 16, 2004 #6
    So that was my mistake! Thanks. That yielded the correct magnitude of the acceleration.
     
  8. Aug 16, 2004 #7
    I must say that I find this explanation lacking ... How can the algebra of inequalities depend upon a foreknowledge of the answer to the problem? Sure, I know that ap is going to be negative, but shouldn't the solution to the equations be what ultimately reveals that? Can anyone think of a different explanation? This is really bugging me.
     
  9. Aug 16, 2004 #8
    Thanks!

    I found your alternative solution using relative velocities to be very elegant, and a lot less time-consuming than my original solution. Thanks for showing me this.
     
  10. Aug 17, 2004 #9

    HallsofIvy

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    In working "applications", you certainly are allowed to use common sense to simplify the problem. IF the problem were simply solving the inequality
    (-29 / ap) < 41.3
    Then one would argue "IF ap is positive, -29/ap< 41.3 is always true (the left side is negative, the right positive) so the solutions set is {x| x> 0}. IF ap is negative, -29> 41.3 sp so p< -29/41.3 and the solution set is {x|x< -29/41.3}. The set of all possible solutions is the union of those two sets".

    There is, of course, a "discontinuity" between those two sets since the inequality itself is undefined at ap= 0.
     
  11. Aug 17, 2004 #10
    Suppose that x denotes the position of the passenge train at time t (the origin of this time t is 0.4seconds after the driver sees the freight train). So,
    [tex]
    \frac{d^2x}{dt^2} = -a (a>0)
    [/tex]

    If however, you wrote
    [tex]
    \frac{d^2x}{dt^2} = a
    [/tex]

    instead of prefixing a minus sign yourself (which isn't bad practice at all), you would get an inequality like the one HallsofIvy has mentioned (and the one you have got in your solution). However, accelerations are allowed to be negative--in the case that there is deceleration. All kinematic variables (except time of course, but usually time is not considered as a kinematic variable) are allowed to be negative and once you adopt increases as positive along the direction of motion, you will get the sign as negative for a decelerating value of acceleration (or more precisely, negative acceleration implies acceleration in the direction opposite to the direction of motion).

    You really don't need to solve the inequality as you would do in mathematics since it is assumed here that acceleration will not be zero (for if it were, you cannot start off with an equation like [tex]v^2 = u^2 + 2aS[/tex] to get [tex]s = \frac{v^2 - u^2}{2a}[/tex] since division by zero is not allowed in the first place!!). That gives us just one set. From the physical point of view, only the extremum value is required here. The interpretation that "this is the maximum possible deceleration that the passenger train must have in order to marginally avoid a collision with the freight train" clearly helps then.

    As a general rule, you should define the sign convention through the initial values, right at the start of a problem such as this one, for that will help you interpret the results obtained at every stage. This is particularly helpful while setting up and solving differential equations.
     
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