Welcome to Netrider ... Connecting Riders!

Interested in talking motorbikes with a terrific community of riders?
Signup (it's quick and free) to join the discussions and access the full suite of tools and information that Netrider has to offer.

Highschool physics cops and robbers question :)

Discussion in 'The Pub' started by robsalvv, Nov 6, 2008.

  1. This is apparently a real world situation... I've run my own calcs but wouldn't mind some independent verification... so fire up your spreadsheets... here's the set up:

    A moving police car detects an oncoming vehicle travelling at 150km/h at point A. The police vehicle slows, stops and conducts a U turn to commence pursuit. The speeding vehicle is intercepted at point B some way down the road.

    - Police vehicle takes 20 seconds to slow, stop, conduct a U turn and return to point A
    - Assume the speeding vehicle begins accelerating uniformly immediately after detection to 211km/h over 8 seconds and then maintains that speed until interception.

    1. How far has the speeding vehicle travelled in the time it takes for the police vehicle to return to the detection point?
    2. Using point A as a reference point, what would be the average speed of the police car if the intercept point B is respectively a) 4km b)5km c)6km from point A.

    Look forward to the answers... and some possible follow up discussion. :)
  2. *grumps* But I'm busy today!



    ... oh alright. give me a bit. :p

    Edit: Question - what is the initial speed of the policecar?

    2nd Edit: Eh, don't worry about it, shouldn't change the results that significantly. :p
  3. I'm no physics teacher, but I would fail you if you used a spreadsheet. :LOL:
  4. HA!!! Sure you have.......or maybe you just want Spots to do it for you. :grin:
  5. Planning on running from the cops eh? :p
  6. 5 years of engineering uni taught me one thing...
    the answer is always 7...
    if not google will have it or the guy over there with the USB stick... :grin: :grin: :grin: :grin: :grin: :grin: :grin:
  7. Kinematics is not my strong point... Give me solid state physics any day over it...
  8. Devo, fair 'nuff if it was an actual highschool question and only that scenario was being considered. This is apparently a real world scenario and I need to be able to change the parameters instantly.

    So YES I do have my spreadsheet. It even has some extra bits to factor in a delay in initial acceleration during those first 20 seconds and another bit to work out predicted pursuit average speed and distance based purely on an estimated time to intercept.

    The next complication is to add in a part that will factor in acceleration at any point in the pursuit, not just in the first 20secs.

    Anyway, I have my answers for the scenario described in the OP. If another physics head works out the same answers independently then we're either both right, or both wrong.

    I used the basic kinematic equations for part 1 and part two is fairly basic distance over time stuff.

    Spots, I look forward to your answers. :)

    Brownyy, unfortunately you fail!!! LMAO
  9. Alright, here we go.

    Edit: Shit, oops. Used kph for the acceleration rate. :p I fixed.

    Question A:
    Method - Find the distance covered by the motorcycle during its acceleration period (S1), assuming uniform acceleration (as per Rob), then find the distance covered at 211kph for the remaining 12 seconds of U-turn time (S2).

    S1 = ut + 0.5 a t^2 where t = 8 seconds
    150kph = 41.66667m/s
    211kph = 58.6111m/s

    a = 2.118m/s^2 ((58.611 - 41.666)/8 seconds)

    S1 = (41.6667m/s * 8s) + (0.5 * 2.118m/s^2 * 8s * 8s)
    S1 = 401.111m covered while accelerating to new speed

    S2 = ut
    S2 = 58.61111m/s * 12 seconds = 703.3333m

    Total distance covered during U-turn = 1104.44m

    Question 2:
    Method: We know the distance travelled by the motorcycle during its acceleration period and that the velocity of the motorcycle is constant for the remainder. We use this knowledge to find out how long it took the bike to cover the respective distances, and from that determine the average speed of the police car.

    4000m - 401.111 = 3598.889m at constant velocity
    3422.6667 / 58.611 m/s = 61.402 seconds at constant velocity
    61.402s + 8 seconds = 69.40284 seconds until police car intercepts bike.

    4000m / 69.403 = 57.63m/s avg, or 207.48kph

    Makes sense; the bike wasted a lot of time at slower velocities, so the average total speed across the distance is slower than the bike's peak speed.

    5000 - 401.11= 4598.89m @ constant V
    78.464 seconds @ constant V
    86.464 seconds to intercept.

    Average speed: 57.82m/s or 208.18kph

    6000-401.111= 5598.89m @ constant V
    95.526 seconds @ constant V
    103.53 seconds to intercept.

    Average speed: 57.95m/s or 208.64kph

    Edit: This is the bike's average speed. Oops. For policecar avg speed, see page 2 of thread.
  10. no i don't, i got a bit of paper :grin: :grin: :grin:
    worked for the rest of my class too! :LOL: :LOL: :LOL:
  11. If I read Q1 correct the answer is about 1110 metres.
    Still working on Q2 :roll:

    Can we go back to discussing weighting the inside peg? :LOL:
  12. doh, i really should read the question right
  13. Turn 150 km/h into 41.7 m/s and 211 km/h into 58.6 m/s just to keep it all SI.

    Then the car does 58.6 x 12 = 703 m in the 12s after it reaches its top speed.

    Its acceleration during the 8s is 58.6-41.7/8 = 2.11 m/s^2

    and (using s = ut + 0.5 at^2) it does 401.1 m during the 8s, for a total distance from detection 'til the cop car returns to Point A of 1104.1 m.

    So it has an 1104 m head start and is doing a constant 58.6 m/s.

    Here's where a spreadsheet comes in handy to avoid having to manually do the same set of calculations 3 times for parts a, b and c of the question.

    The car being pursued has to do 4000-1104 m at 58.6 m/s: that will take 49.4 s. The cop car has to do the whole 4000 m in that same time, so its average speed will be 4000/49.4 = 80.97 m/s or 291.5 km/h.

    Using the same calculations the 5 km question will require a slightly lower maximum speed of 270.7 km/h and the 6 km will require 258.7.
  14. Spots got through the calculation while I was doing it (hey, I had to use the calculator on the computer!), but unaccountably got a small error in the initial distance...
  15. Ginji, try again.

    Acceleration occurs in the first 8 seconds and then it's 211km/h for the remaining 12seconds.

    And the acceleration I've worked out using V''=V'+a*t is 2.12m/s/s

    That means

    a= (V''-V')/t

    211km/h = 58.7m/s
    150km/h = 41.7 m/s
    Delta = 16.9m/s over 8seconds which equals 2.12m/s/s

    And the other bit you've failed to realise is that the speeding vehicle is ALREADY a number of meters down the road when the cop is back at pointA... sooo whatever time it take the speeding vehicle to get to the intercept point is ALL THE TIME the cop has to cover the ENTIRE PURSUIT distance from ptA to ptB.

    Goodluck on your physics exam.
  16. Yea, I read the question wrong, I thought it was the cop car accelerating to 211km/h.
  17. Fark me!!! how many responses... ok... I'm concerned about the acceleration. How can I get 2 and other 7???
  18. Yeah, I realised that I'd calculated acceleration in kph/s rather than m/s^2 by picking the wrong cells.

    The hazards of using Excel rather than Mathcad. :LOL:

    Edited my solution with the correct figures.
  19. We still have different answers for the last bit, Spots.
  20. PS No spreadsheets were used in the production of my answers, just a dodgy Mac non-scientific calculator