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Physics Question Help

  1. Oct 11, 2003 #1
    The question is:
    A cheetah cah hit 105 km/h in 2s, and maintain this speed for 15 s, while an antelope can reach 90km/h in 2s and sustain this for a long time.. if they are both separated by 100m and the antelope reacts 0.5s after the cheetah starts running, will they meet? and if not, when?

    Basically what I did was I calculated how far each would go in the 2s..

    for the cheetah:
    I basically used the formula X = 0.5at^2
    and came up with 58.3m

    For the antelope: I came up with 150 m.

    (Cheetah is the point of origin).

    Then, I basically added 0.5s to the constant velocity of 105 km/h (or 29.1m/s) and found out that in 2.5 s, the cheetah travelled 72.9 m.

    Thus, I subtracted the two distances, of 150m - 72.9m and this results in 77m.

    Then I set the point of reference so that the antelope is at rest and the cheetah is moving at 4.17m/s (relative to the antelope).

    Thus, I calculated that the time would take 18.48 s

    Therefore, they would not meet since the cheetah can only run for 15s.

    However, it asks how far they are apart.
    Therefore, I assume that since the constant velocity of the cheetah is greater than the antelope, at exactly 15s, the cheetah will be the closest.

    Therefore, at this point I'm stuck on what to do..

    I subtracted 2 from 15 (so 13) multiplied by 4 m/s but that didn't give me the right answer...

    (of 26.6m)

    Any help is appreciated! :)
  2. jcsd
  3. Oct 12, 2003 #2
    I think these are both false.
    For the cheetah:
    a = 105km/h / 2s = 105km / 3600 s / 2s = 14.6 m/s2.
    d = .5at2 = .5 * 14.6 m/s2 * 4 s2= 29.2m

    The rest of your approach looks OK to me.
    Well, calculate how far each of them has gone after 17 seconds. (That would have been the first thing I'd have done...)
  4. Oct 12, 2003 #3
    Oops, Thanks for the help!
  5. Oct 13, 2003 #4


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    First point: does the cheetah catch the antelope before the first 2 seconds are up? Since the cheetah goes from 0 to 105 km/hr = 29.2 m/s in two seconds, assuming constant acceleration, it has an average speed of 29.17/2= 14.6 m/s and so, during those first two seconds, its distance covered in time t is 14.6 t.
    The antelope goes from 0 to 90 km/hr= 25 m/s in 2 seconds so, assuming constant acceleration, has an average speed of 25/2= 12.5 m/s. Its distance covered in time t is 12.5t (0< t< 2). Since he had a 100 m "head start" distance from the cheetah's starting point is
    12.5t+ 100 for the antelope.

    The cheetah will catch the antelope in the first 2 seconds if
    14.6t= 12.5t+ 100 for t< 2. That's the same as 2.1t= 100 or t= 47.6 seconds. No, the cheetah doesn't come any where near catching the antelope in the first 2 seconds! In fact, we can calculate that the distance the cheetah covers in those first two seconds is 14.6(2)= 29.2 meters while the antelope covers 12.5(2)= 25 meters. Since the antelope had a 100 m lead, after 2 seconds, the antelope is still 100+ 25- 29.2= 95.8 meters ahead.

    After this, the cheetah runs at a constant 29.2 m/s so its distance covered is 29.2 t (t is now seconds since the initial 2 second time ended). The antelope runs at a constant 25 m/s so its distance covered is 25t. The cheetah is gaining 29.2- 25= 4.2 m every second. It will catch the antelope if it can gain the entire 95.8 m left- that is, when 4.2 t= 95.8 which is t= 22.8 seconds.

    Unfortunately for the cheetah (and fortunately for the antelope), the cheetah can only sustain that speed for 15 seconds. 22.8 seconds is too long. The cheetah does not catch the antelope.

    4.2(15)= 63 meters. When the cheetah can no longer maintain that pace, the antelope is still 95.8- 63= 32.8 meters behind. It wasn't even close!

    By the way, you question was "will they meet? and if not, when?" which doesn't make sense. They will never meet! I presume you mean "if yes, when".
  6. Oct 13, 2003 #5
    Thank you for your help!
    Sorry for the confusion. I meant to say that if they don't reach each other, how close do they get?

    However, I don't think the answer in the book is correct.

    The book says that they are the closest at 26.6 m, but based on my numerous trials and verifying my answers with yours, it doesn't seem logical.

    At anyrate, thanks again!
  7. Oct 13, 2003 #6
    OK, one more time. Let's calculate both positions after 17 seconds. WRT to cheetah start position.

    cheetah: d = 14.6m/s * 2s + 29.2m/s * 15s = 467.2m
    antelope: d = 100m + 12.5m/s * 2s + 25m/s * 14.5s = 487.5m

    difference: 487.5m - 467.2 m = 20.3m

    Why is HallsofIvy wrong? Because he did this:
    antelope: d = 100m + 12.5m/s * 2s + 25m/s * 15s = 500m
    500m - 467.2m = 32.8m
    He omitted the half-second delay.

    Why is the book wrong? - Because they did this:
    antelope: d = 100m + 12.5m/s * 1.5s + 25m/s * 15s = 493.8m
    493.8m - 467.2m = 26.6m
    They assumed the antelope accelerates for 1.5 seconds, then runs for 15 seconds.

    Edit: color
    Last edited: Oct 13, 2003
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