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Homework Help: Relative motion -cheetah and gazelle

  1. Apr 17, 2009 #1
    1. The problem statement, all variables and given/known data
    A cheetah can run at approximately 100km/hr and a gazelle at 80km/hr. If both animals are running at full speed, with the gazelle 70m ahead, how long before the cheetah catches its prey?

    2. Relevant equations
    s=vt (no acceleration)
    v = km/hr / 3.6 = m/s

    3. The attempt at a solution
    I understand that this is relative motion. I've calculated that it takes the cheetah 2.52 s to run 70m and that during that 2.52 sec the gazelle would run an additional 55.9 m so that the total distance that the cheetah has to run is 125.9 meters. I've calculated that the time it takes for the cheetah to run this distance at constant speed is 4.53 s. I've redone this question about 10 times and can't figure out what I'm doing wrong. Can anyone please shed some light on what is probably a ridiculously easy question. Thank you.
  2. jcsd
  3. Apr 17, 2009 #2


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    Homework Helper

    Make things simpler on yourself. Put yourself inside the head of the cheetah.

    The cheetah sees dinner just 70 m away, and coming closer at the rate of 20 km/h.

    Looked at another way, a student is waiting for a Pop Tart being delivered on a conveyor belt going at 20 km/h. The cafeteria worker is putting it on the conveyor from 70 m away. How long?

    So how long does it take for that meal going at 20 km/h to travel 70m?
  4. Apr 17, 2009 #3
    I've been looking at relative motion equations in 2d and having no problem but when I saw this one my brain was certain that somehow it had to be super complicated. Its not though. Thanks, is all makes sense now.
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