Anonymous
Anonymous asked in Science & MathematicsPhysics · 9 months ago

Physics Question: A vehicle accelerates from rest to 13.5 m/s in 12 seconds. How far did the vehicle travel? What formula(s) do you use?

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  • MyRank
    Lv 6
    8 months ago

    Given,

    Initial speed (u) = 0m/sec

    Final speed (v) = 13.5m/sec

    Time (t) = 12sec

    Distance (d) =?

    Using kinematic relation:-

    v = u + at

    13.5m/sec = 0m/sec + a x 12

    13.5m/sec = a x 12

    Acceleration (a) = 13.5/12 = 1.125m/sec²

    Now, v² = u²+2ad

    (13.5)² = (0)² + 2 x a x d

    182.25 = 2 x 1.125 x d

    Distance (d) = 182.25 / 2.25 = 81m

    ∴ Distance (d) = 81m.

  • Anonymous
    9 months ago

    Vavg = Vfinal / 2 = 13.5m/s / 2 = 6.75 m/s

    and then

    distance d = Vavg * t = 6.75m/s * 12s = 81 m ◄

    OR you could find the acceleration

    a = Vfinal / t

    and then

    d = ½ * a * t²

    with, of course, the same result.

    Hope this helps!

  • Anonymous
    9 months ago

    stretched distance d = (Vf+Vi)/2*t = (13.5+0)*12/2 = 13.5*6 = 27*3 = 81.0 m

    or

    acceleration a = ΔV/Δt = (13.5-0) / (12-0) = 27/24 = 9/8 of m/sec^2

    stretched distance d = a/2*t^2 = 9/16*12^2 = 144*9/16 = 81.0 m

    or

    stretched distance d = (Vf^2-Vi^2) / 2a = (27^2/4-0)*8/18 = 27^2*2/18 = 81.0 m

  • CRR
    Lv 7
    9 months ago

    V=at

    D=at^2/2

    (Assuming constant acceleration)

    You know V and t so you can find a and D.

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