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Motion with Constant Acceleration

  1. Sep 8, 2008 #1
    Please I need help with these two problems...the answer is in the back of the book but as much as I've tried I can't find the procedure to do them:cry:

    I really appreciate your help...thank you so much beforehand!

    1) The operation manual of a passenger automobile states that the stopping distance is 50 m when the brakes are fully applied at 96 km/h. What is the magnitud of decelaration? What is the stopping time?

    Answer: -7.1 m/s^2

    2) The front end of an automobile has been designed so that upon impact it progressively crumples by as much as 0.70 m. Suppose that the automobile crashes into a solid brick wall at 80 km/h. During the collision the passenger compartment deccelerates over a distance of 0.70 m. Assume that the decceleration is constant. What is the magnitud of the decceleration?

    Answer: -350 m/s^2
     
  2. jcsd
  3. Sep 8, 2008 #2
    hey! I could help u......for problem no.1
    tried this equation it may help you! let:
    V^2=v^2+2ad

    where:
    V=final velocity
    v=initial velocity
    a=acceleration
    d= displacement (in unit of meter)

    but; first you must
    convert km/hr to m/s
    and v=initial velocity will be zero because
    it start at rest..... (automobile)
     
  4. Sep 8, 2008 #3
    this is easy...
    ok now it says stopping distance is 50m
    Since brakes are applied, final velocity is 0. Initial Velocity is 96 km/h.

    so S=50m, u= 96km/h = 26.67m/s, v = 0.

    so v^2 = u^2 + 2aS

    so a = v^2 - u^2/ 2s = 0 - (26.67)^2/2X50 = -711.28/100 = - 7.112 m/s2
     
  5. Sep 8, 2008 #4
    Oh sorry forgot about time....since v = u + at.....t = v-u/a

    so t = 0 - 26.67/-7.1 = -26.67/-7.1 = 3.76 sec
     
  6. Sep 8, 2008 #5
    great....
     
  7. Sep 8, 2008 #6
    Thank you SO much you guys :biggrin:!

    Homeworkboy I have a question...what does "S" and "u" stand for?

    thank you again lol
     
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