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Homework Help: Kinematics motion-police's radars(answer check)

  1. Dec 20, 2013 #1
    1. The problem statement, all variables and given/known data

    A police radar's effective range is 1km and your radar detector's range is 1.9km. You're going at 110kmh^-1 in a 70kmh^-1 zone when the detector beeps. At what rate must negatively accelerate to avoid a speeding ticket?

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Dec 20, 2013
  2. jcsd
  3. Dec 20, 2013 #2


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    Your answer looks correct. But a couple of points: firstly, you should be careful about your sign conventions. It appears that you took the police radar gun's position as being the origin (this should be stated). Distances going outward from the police position toward you are increasing in a positive direction. So you start out at +1.9km (when your detector goes off), and you're heading in a negative y direction toward the police position. Your aim is to decelerate to 70km/h before you hit a position of +1.0km, when the police radar would've acquired your speed.

    This convention is fine, but in this case, remember that your velocities are actually all negative. It doesn't matter when you square them, of course, but you should still write them correctly within the parantheses. Also, note that the deceleration you calculate will be an acceleration directed along the positive direction (i.e. a "push" away from the police cruiser, opposite to your direction of your car's motion), and hence will come out positive, which explains why the "a" value you calculated is positive. Nothing wrong with this, but it's worth noting why it came out positive when the usual convention is to have a negative value signifying a deceleration.

    A slightly more intuitive approach would be to let your car's position at the point of your radar detector beeping be the origin. Your ordinate increases in the direction of travel, and your aim is to decelerate from +110km/h to +70km/h before you hit +0.9km (which is where the police radar can pick up your speed). This way, you get a conventional negative answer for "a".

    One more thing: about your units. It really would be better to convert everything to base SI units before doing the calculation. The numbers work out fine in what you did, but some of the units are wrong. For instance when you square the velocities and subtract, the result should be expressed in ##km^2h^{-2}## and not ##kmh^{-1}## as you did. Also, the denominator should have a unit, i.e. ##(3600sh^{-1})^2 = (3600)^2s^2h^{-2}##. It's also clear that you didn't include a factor of 1000 in that expression, even though you did actually multiply by it before getting the final answer. The expression should really be:

    ##\displaystyle \frac{\frac{4900 km^2h^{-2} - 12100km^2h^{-2}}{(2)(-0.9km)}}{10^{-3}(km)m^{-1}(3600sh^{-1})^2}##, which would give you the correct answer. But this is very unwieldy, and you're better off just converting to base SI units from the get-go.
    Last edited: Dec 20, 2013
  4. Dec 20, 2013 #3


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    looks good (Your answer was actually .0003 km/s^2, which you apparently converted behind the scenes to 0.3m/s^2)
    Edit: ..and with signage/unit errors as noted by Curious.
    Last edited: Dec 20, 2013
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