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9 July, 09:52

A car of weight 1.8 x 10^4 N is accelerated uniformly from rest by a force of 5.0 x 10^3 N east for 6s. Calculate the following for the car the displacement from its original position.

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Answers (2)
  1. 9 July, 11:30
    0
    50.04 m

    Explanation:

    Weight of car = 1.8 x 10^4 N

    u = 0

    F = 5 x 10^3 N east

    Let d be the displacement in the position of car in 6 s.

    Mass of car, m = Weight / g = 1.8 x 10^4 / 10 = 1.8 x 10^3 kg

    Force = mass of car x acceleration of car

    a = F / m = 5 x 10^3 / (1.8 x 10^3) = 2.78 m/s^2

    Use third equation of motion

    s = u t + 1/2 a t^2

    d = 0 + 1/2 x 2.78 x 6 x 6

    d = 50.04 m
  2. 9 July, 11:40
    0
    Displacement = 50.04 m

    Explanation:

    Given,

    Weight of car, W = 1.8 * 10⁴ N

    Initial Velocity, u = 0

    Force, F = 5 * 10³ N (at the East direction)

    Time, t = 6s

    Let 'd' be the displacement in the position of car in 6 seconds.

    Mass of car, m = Weight/Gravity

    m = 1.8 * 10⁴ N / 10 m/s²

    m = 1.8 * 10³ kg

    Now,

    Force of car = Mass of car * acceleration of car

    So, Acceleration = Force/Mass

    Acceleration = 5 * 10³ N/1.8 * 10³ kg

    Acceleration = 2.78 m/s²

    By using the formula of second equation of motion.

    s = ut + 1/2 at²

    where, u = 0

    So,

    s = 1/2 at²

    s = 1/2 * 2.78 * 6²

    s = 1/2 * 2.78 * 36

    s = 50.04 m

    Here, s is 'd'. 's' and 'd' both are displacement.
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