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Highway Patrol - Argh, looks so simple!

  1. Sep 30, 2003 #1
    Suppose you are driving on the interstate at 85 mph when you see a highway patrol car parked up ahead. You apply your brakes, decelerating at a constant rate of 4 mph/sec until you're reduced your speed to 65 mph. You then maintain a constant speed of 65 mph. You pass the patrol car 15 seconds after you initially applied the brakes and are relieved that he doesn't pull out after you. How far away was the highway patrol car when you first saw it?

    And it wants the answers in meters...
    This is what I tried doing so far, but to no avail.
    v0=85mph => .0236 m/sec
    vf=65mph => .0181 m/sec

    Okay. I've tried this equation x-x0=(1/2)(v0+v)*t and solved for x with an answer of .31275 m, but that doesn't work. :-(

    I've tried vf^2=v0^2+2*a(xf-x0) and solve for displacement, but that answer doesn't work either.

    Maybe I'm interpreting the initial values wrong or something. Any insight appreciated.
  2. jcsd
  3. Sep 30, 2003 #2
    Your conversions from miles per hour to meters per second are incorrect.

    85mi/hr = 37.99m/s
    65mi/hr = 29.06m/s

    And you can't use -4mph/s as an acceleration because the units do not work. You have to convert everything to either hours or seconds, you can't mix them. Your units have to be either mph/h or mps/s.

    Try changing those values and see where you get.
  4. Oct 1, 2003 #3


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    ThatOneGuy (I thought you were the other guy!): I misunderstood also: RockDog means MILES per second, not meters per second.

    His conversion for that is correct: 85 mph/3600 s/hr= 0.0236 miles per second. 65 mph/3600 s/h= 0.0180 miles per second.

    The error is in not accounting for the deceleration and constant speed parts separately.

    The driver decelerates from 85 to 65, a reduction of 20 mph, at a deceleration of 4 mph per second and so decelerates for 5 seconds.
    Since the deceleration is constant, you can treat the entire 5 seconds as if it were at the "average" speed, 75 mph (that's the point of RockDog's equation "x-x0=(1/2)(v0+v)*t "). Of course, you could also calculate d= v0t+ (1/2)a t2.
    Using the "average" speed, 75 mph= (75)/3600= 0.0208 miles per second so in 5 seconds he went (0.0208)(5)= 0.104 miles.
    Using the quadratic formula, d= (0.0236)(5)- (1/2)(0.0011)(25)
    = 0.104 miles (notice that I converted "4 mph per second" to "0.0011 miles per second per second).

    Either way, we arrive at 0.104 miles traveled during the deceleration.

    Now the car continues at 65 mph (= 0.0180 miles per second) for 15 seconds which would give (0.0180)(15)= 0.27 miles.

    The car traveled a total of 0.104+ 0.27= 0.374 miles.
  5. Oct 1, 2003 #4
    Sup guys. I figured something out, HallsofIvy, you messed up on a tiny little part. :-)

    Since the car decelerated for 5 seconds, that meant it would have 10 seconds left cruising at a constant speed of 65mph. Thus, the last stretch of distance would be 65 mph (= 0.0180 miles per second) for 10 seconds = .180 miles.

    Then just convert total miles to meters.
    Last edited: Oct 1, 2003
  6. Oct 1, 2003 #5


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    Right: I misread the problem. I thought it said 15 seconds at constant speed instead of 15 seconds total time.

    But why convert miles to meters? The problem gives everything in terms of miles- give the answer that way.
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