Ask Me Help Desk

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)
-   Math & Sciences (https://www.askmehelpdesk.com/forumdisplay.php?f=402)
-   -   Physics of Motion (https://www.askmehelpdesk.com/showthread.php?t=64925)

  • Feb 21, 2007, 09:14 AM
    Evil dead
    Physics of Motion
    Hi.

    I have 2 animals a cheetah and an antelope.

    From previous question set for me I have been given the following information:

    • The cheetah accelerates from rest (0ms-1) to 20ms-1 in 2 seconds
      It has a top speed of 30ms-1
      It can only maintain this speed for 450m before it has to rest
      An antelope however has a top speed of 22ms-1


    I have worked out:

    • The cheetahs average acceleration is 10ms-1
      From rest (0ms-1) it takes the cheetah 3 seconds to get to it's top speed
      The cheetah travels 45m while accelerating to it's top speed.
      It takes 15 seconds traveling 450m from its top speed (30ms-1) before it has to rest
      The cheetah can run a combined duration of 18 seconds (it's time to accelerate 3 seconds + it's time from it's top speed before it as to rest, having travelled 450m from it's top speed 15 seconds)
      The antelope has the same rate of acceleration as the Cheetah, therefore it takes 2.2 seconds for it to accelerate from rest to 22ms-1 (it's top speed) - that is what we can assume, if it has an average acceleration of 10ms-1.


    The question: If an antelope starts from rest and accelerates to it's top speed at the same rate as the cheetah, how far will it travel in the time obtained in your answer to d (18 seconds)?
  • Feb 21, 2007, 09:53 AM
    Capuchin
    Okay, for the accelerating portion, your antelope has average speed of 22/2 = 11m/s

    so he travels 11*2.2 = 24.2 meters

    (you get the same result using s = ut + 1/2 at^2, which is a more sensible way of looking at it.)

    then for the rest of it (18-2.2 = 15.8s) he is travelling at 22m/s

    so this is 22*15.8 = 347.6m

    total distance ran is 347.6+24.2 = 371.8 m

    I think this is right :)
  • Feb 21, 2007, 10:16 AM
    Evil dead
    Its is right (solutions says so).

    I knew before that the total distance travelled is the distance for acceleration + the distance from top speed to rest.

    I don't understand this part however: your antelope has average speed of 22/2 = 11m/s

    The question says that the antelope accelerates the same rate as the cheetah, which means it's average acceleration is 10mph. So in order to get from rest to 22, it needs 2.2 seconds.

    Also: your antelope has average speed of 22/2 = 11m/s
    Where did you get the 2 from?
  • Feb 21, 2007, 10:17 AM
    Capuchin
    if acceleration is constant, then the average speed is start speed (0) plus final speed (22) over 2 (number of speeds to average).

    This is an awkward way of doing it. I suggest you learn to use the equation I gave underneath. It's far less confusing :). This is just the way that makes sense in my head.

  • All times are GMT -7. The time now is 05:17 AM.