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A certain car is capable of accelerating at a uniform rate of 0.93 m/s2. What is the magnitude of the car’s displacement as it accelerates uniformly from a speed of 86 km/h to one of 98 km/h?
Answer in units of m

  • physics -

    vf^2=vi^2+2ad solve for d.

  • physics -

    be certain to change km/hr to m/s...

  • physics -

    v = Vi + a t

    Vi = 86 km/h * 1000 m/km * 1 h/3600 s
    = 23.9 m/s

    Vf = 98 km/h *10/36 = 27.2 m/s

    hard way:
    27.2 = 23.9 + .93 t
    so
    t = 3.55 s

    x = Xi + Vi t + (1/2) a t^2
    x-Xi = displacement = 23.9(3.55)+.465(3.55)^2
    = 90.7 m

    easy way:
    average speed = (23.9+27.2)/2 = 25.55
    change in speed/t = .93 = 3.3/t
    t = 3.55 sure enough
    distance = 25.55*3.55 = 90.7 m

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