Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Crash with the police car /suvat problem

  1. Sep 17, 2011 #1


    User Avatar

    1. The problem statement, all variables and given/known data

    You are arguing over a cell phone while trailing an unmarked police car by 25.0 m; both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.0 s (long enough for you to look at the phone and yell,“I won't do that!”). At the beginning of that 2.0 s, the police officer begins braking suddenly at 5.00 m/s2. (a) What is the separation between the two cars when your attention finally returns? Suppose that you take another 0.400 s to realize your danger and begin braking. (b) If you too brake at 5.00 m/s2, what is your speed (in km/h) when you hit the police car?

    2. Relevant equations

    [tex] v^2 =u^2 +2as, s=ut +\frac{1}{2} at^2 [/tex]

    3. The attempt at a solution

    I solved the first part. I get the wrong results for part b) because I think I'm making calculation mistakes. The way how I tried to solve it is by making the distances traveled equal.

    From [tex] s=ut +\frac{1}{2} at^2 [/tex] where [tex] a=\frac{-5m}{s^2} [/tex]$ we have that since I start breaking [tex]2.4s[/tex] later than the police car, I must travel a distance [tex]s=u(t-2.4) +\frac{1}{2}at^2 [/tex] whilst the police car travels a distance [tex]s'=ut +\frac{1}{2}at^2 [/tex]. Since we crash, we must have [tex] s=s' [/tex]... this way finding [tex] t [/tex] and [tex] s [/tex], and from there using [tex] v^2 =u^2 +2as [/tex] we find [tex] v [/tex]

    P.S. The LaTeX looks weird.
  2. jcsd
  3. Sep 17, 2011 #2


    User Avatar
    Homework Helper


    For the first 2.4 seconds, you continued at constant speed while the Police car braked [note: it didn't break - it only did that after you hit it].

    Using your formulae [or a more suitable one] you could calculate the distance you have both travelled, and thus how far you are then behind the PC.
    You can also calculate how much faster than the police car you are travelling at that time.
    From that time on, both you and the PC have the same acceleration, so you will continue to be travelling that same amount faster than the police car, until it, then eventually you, stops - if you haven't already collided.

    Let's invent some number - apologies if any of them are actually the correct ones.

    Lets assume the initial speed is 30 m/s [it is actually a bit faster]
    In 2.4 seconds you will travel 72m
    In 2.4 seconds the police car will have slowed to 18 m/s, so average speed is 24 m/s so it has progressed 57.6 m
    You have caught up 14.4 m so are now only 10.6m behind.
    You are travelling 12 m/s faster than the police car at all times while braking [you each have the same acceleration] so it takes a fraction of a second to catch the Police Car, your speed will have reduced by only perhaps 4 m/s, so you will be travelling quite fast.
    Note that your closing speed will only be 12 m/s, so the impact damage won't be too great, but the "spinning off into the scenery" that both you and the Police Car are about to begin could be exciting

    NOTE: You might want to rework the problem without the phone, and just include the 0.400 seconds reaction time to see if you stood a chance of stopping before hitting the Police Car anyway!!!!! 25m is about 3 car lengths, and on a 110km/h freeway is probably too close to be following anyway!!!!!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook