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Speed calculation/falling bodies theory

  1. May 25, 2007 #1
    I need a formula(s)!!!
    I have a vehicle(semi) traveling on a 2% downgrade with a begnining speed of 60mph, no acceleration or braking on the downgrade. Adj "f" is .73 if that is needed. I need to be able to calculate the vehicles increase in speed in both time and distance increments. Doesn't matter what the incriments are, 20ft, 100ft or every 10sec of travel, it doesnt' matter.
  2. jcsd
  3. May 25, 2007 #2
    I thought you said "no acceleration or braking on the downgrade," so how is it to increase in speed?
  4. May 25, 2007 #3
    I guess I'm assuming the vehicle would speed up going down hill. I'm also going to have to assume it was in neutral for now.
  5. May 26, 2007 #4
    I guess I should clarify what I'm trying to ask. By saying no acceleration I meant that the driver will be letting off the gas and not using the accelerator. I didn't explain what I meant very well.
  6. May 26, 2007 #5


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    The 2% downgrade gives you 2% drop per horizontal distance traveled which may be close enough to 2% drop per road distance so as to treat it as the same. It also gives you a 2% of the acceleration due to gravity. I'm going to work with metric units and then convert using 8km = 5mi.

    For the time increments acceleration will be constant, 2% of 9.8 meters per second:
    [tex]a = 0.02 \times 9.8 =0.196\frac{m}{s^2}[/tex]
    and just convert it to the units you want i.e. miles per hour per second.
    [tex] a = .196\frac{m}{s^2} = .196\times 3600 \frac{m}{s\cdot hr}
    =0.196\times 3.6 \frac{km}{s\cdot hr} = 0.7056\frac{km}{s\cdot hr}[/tex]
    now in terms of miles:
    [tex] a = \frac{5\times .7056}{8} =0.441\frac{mph}{s}[/tex]
    So the speed will increase .441mph every second, 4.41mph every ten seconds, 26.46 mph every minute and so on.

    The distance formula won't be as easy. You have to use work and energy. For every meter you drop the work done per unit mass is:
    [tex] \Delta W/M=0.2\times 9.8=0.192 \frac{m^2}{s^2}[/tex]
    This results in a corresponding increase in kinetic energy per unit mass which is a function of the velocity:
    [tex] KE/M = 0.5 V^2[/tex]
    Given the truck travels initially at velocity [tex]V_0[/tex] the change in kinetic energy will be:
    [tex] \Delta KE/M = 0.5(V^2-V_0^2) = 0.182 x[/tex]
    where x is the distance traveled in meters.
    You then get:
    [tex] V = \sqrt{.364 x+V^2_0} [/tex]
    where speed is in meters per second and distance in meters.
    converting speed from m/s to mph I get
    1 m/s = 3600 m/hr = 3.6 km/h = 5*3.6/8 =2.25mph

    [tex] V = 2.25\sqrt{.364 x + \left(\frac{V_0}{2.25}\right)^2[/tex]
    Now speeds are in mph and distance traveled is still in meters.
    For 60mph you get:

    [tex] V = 2.25\sqrt{.364 x +711.1111m}[/tex]

    I'll leave it to you to convert x from meters to units you like. You see it isn't a simple matter of picking up a certain fixed speed per distance traveled because the faster you move the less time you spend accelerating across a given distance. The velocity dependent term inside the square root of the formula is just the distance needed to get from zero up to the initial speed of 60mph.

    James Baugh
    Last edited: May 26, 2007
  7. May 26, 2007 #6
    Outstanding! Thank you so much.
    Aysha Vom Zettlebauch
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