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Simple Kinematic Question

  1. Feb 22, 2004 #1
    I have a pretty simple highschool physics question and hope anyone here can answer it.

    Question - A space shuttle is coming in for a landing and is approaching the runway with a speed of 450km/h and at an range of 40 degrees below the horizontal. It smoothly changes direction so that when it hits the runway it is traveling horizontally with a velocity of 180km/h. If it takes 45 seconds to make this change, what average acceleration is the shuttle experiencing?

    PS: I hope one on this forum can answer this question.
  2. jcsd
  3. Feb 22, 2004 #2
    i might be able to help

    Im pretty new to physics myself. But I think i will be able to help you, but I am not sure if i am right.

    First i think we have to find the x component of the intial velocity. We find this by noticing the shuttle angle of attack is 40. So with the laws of trig we use "450 cos 40" which equals 344.7m/s for the x component. Then we use the equation (average acelleration= delta V/ delta T) so we take the difference from the intial velocity minus the final velocity, which is 344.7-180= 164.7. then we devide that by the time difference(which is 45) and we get 3.66m/s2. So i think that is the average accelleration, but i am not sure, since i am pretty new at this myself. Hope that helped.
  4. Feb 22, 2004 #3


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    That's pretty good, Swigs, but you do need to take into account the vertical change as well. The shuttle was initially flying at 450 km/h at an angle 40 degrees below the horizontal: its horizontal component of velocity is 450cos(40) and its vertical component is -450 sin(40) (negative because it is downward).

    After the "transition", its speed is 180 km/h and it is moving horizontally: that is, its vertical component of velocity is 0.

    The horizontal component of velocity has change from 450 cos(40) to 180: a change of 450 cos(40)- 180. It did that in 45 seconds. Be careful! Those speeds are given in "km/h". You need to decide if you want to change hours to seconds in them or change seconds to hours in that "45 seconds" (Since there are 60 seconds per minute and 60 minutes per hour, there are 60*60= 3600 seconds per hour. That 45 seconds is 45/3600= 0.0125 hr. I would be inclined to change 450 km/h to 450/3600 (k/hr)(hr/sec)= 0.125 km/sec. In fact I would be inclined to change to meters: (0.125 km/sec)(1000 m/km)= 125 m/sec.)

    That is, in terms of m/s, the shuttle's horizontal velocity was
    125 cos(40) and changed to 180(1000/3600)km/h= 50 m/s. A total change of 125 cos(40)- 50 m/s which my calculator tells me is 45.76 m/s. Since that took place in 45 seconds, the average acceleration was 45.76/45= 1.017 m/s2.

    The vertical change in speed was from 450 sin(40)km/h (which is 125 sin(40)= 80.3 m/s to 0 in 45 seconds: an average of 80.3/45= 1.79 m/s2 (even larger than the horizontal acceleration!).

    That is, the "average acceleration" vector has components 1.017m/s and 1.79 m/s.

    If you want "average acceleration" as a number, find the "length" of that vector: √(1.0172+ 1.792).
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