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Homework Help: Minimum distance

  1. Sep 1, 2003 #1
    I know I've asked quite a few questions, but I really am making an effort to get as far as I can. The engineer of a train moving along a level track with a velocity of 42.0 m/s sights a freight train at a distance of d ahead of him on the same track moving in the same direction with a velocity of 18.0 m/s. He applies the brakes, giving his train a constant acceleration of -1.4 m/s^2. What is the minium distance d such that there is no collision?

    Here's what I came up with. I used the formula d/delta x = 42.0t+1/2(-1.4)t^2. I derived the formula giving me 42+(-1.4)=18. Solved for t getting 17.143. Plugged t back into the original formula I used and ended up getting 514.288. That isn't the correct distance..

    Here's another question..
    Let's say that you are driving a car that accelerates according to a=B/v where B= 130.0m^2/s^3 is a parameter that is related to the ratio of your car's power to its weight, and c is your car's speed, in m/s. Assume that you are initially traveling with a speed of 11.5 m/s. At t=0 you step on the gas pedal. The car performs a constant-power acceleration until you reach a speed of 23.5 m/s. What is the time interval needed to make this change of speed.

    I solved for a by plugging in the values given for B and v, so a=130.0m^2/s^3/11.5 m/s. I then got a= 11.3m/s^2. From there I went to 11.5+11.3t=23.5. Solved for t and got 1.062 s. 1.062s isn't the correct answer. Was my approach to this problem incorrect?
    Last edited: Sep 1, 2003
  2. jcsd
  3. Sep 2, 2003 #2
    Your first question is equivalent with the following: how far does a stone go up (what height) if one throws it upwards with (42-18)m/s and the gravitational pull is g=1.4m/s2 (about the same as on the moon).
    For the second one I have another advice for you: start thinking the problems "in letters" first and then plug in the numbers. You have a*v=B=constant (as the power remains constant at those 175 horse power it has). you also have a=dv/dt. So I now see a diiferential equation with separable variables v and t. Do you know how to solve this one?
  4. Sep 2, 2003 #3
    Nope I still don't know how to solve it (the 2nd question). How did you figure out the car had 175 horse power? I don't think it matters. Are you saying I should integrate a=dv/dt?
  5. Sep 2, 2003 #4


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    Looks to me like you have done this correctly. you have x= 42.0t+ (1/2)(-1.4)t^2 (that's the distance the train goes in time t. I don't understand the "d/delta" part.) Differentiating (which in my opinion is better than "deriving") gives velocity and, of course, the train can avoid a collision by slowing to 18 m/s: dx/dt= 42.0- 1.4t= 18 (you left the t out in your formula. I hope that was a typo) which gives -1.4t= 18-42= -24 so t= -24/-1.4= 17.1 seconds precisely what you got. Now put that back into the formula for x: 42.0(17.1)+(1/2)(-1.4)(17.1)^2= 718.2-0.7(292.41)= 718.2-204.7= 513.5 m. Now, why do you say that is not the correct distance?

    a= B/v is the same as dv/dt= Bv or dv/v= Bdt. Integrating both sides, ln|v|= Bt+ Const or v(t)= e^(Bt+ Const)= e^(Bt)e^Const= C e^(Bt) where C= e^Const. Since your initial speed is 11.5 m/s,
    v(0)= C e^0= C= 11.5 so v(t)= 11.5 e^(Bt). Your car reaches 21.5 m/s when v(t)= 11.5 e^(Bt)= 21.5 or e^(Bt)= 21.5/11.5= 1.87. Now take the natural logarithm of both sides: Bt= ln(1.87). Since you know B you can now solve for t.
  6. Sep 3, 2003 #5
    I figured the car weighs about 1000kg so power=B*1000Kg.
    I'm saying you should do the following:
    integrate: ∫11.523.5 dv*v= ∫0t' dt'*B
    giving: (1/2)v2|11.523.5 = (23.52-11.52)/2 = B*t

    HallsofIvy: it's a=B/v, not a=Bv. Man, this would be easy if we could use the latex &over
  7. Sep 3, 2003 #6


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    Oops! I switch the formula in mid-sentence!

    You say, first, "you are driving a car that accelerates according to a=B/v" and then later say:"The car performs a constant-power acceleration" and seem to be assuming that means a constant acceleration. You calculate B/v for v= 11.4 m/s and then use that constant acceleration. I am inclined to interpret "constant power acceleration" as meaning that the acceleration follows the law you gave: dv/dt= B/v.

    dv/dt= B/v gives vdv= Bdt so (1/2)v^2= Bt+ C. Your initial speed is 11.4 m/s so (1/2)(11.4)^2= C or C= 64.98. Your speed at any time t is given by v^2= 2Bt+ 129.96. You want to increase to 23.5 m/s so you need to solve 2Bt+ 129.96= 23.5^2= 16889.
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