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18 January, 14:56

A record of travel along a straight path is as follows: (a) Start from rest with constant acceleration of 3.9 m/s 2 for 15.3 s; (b) Constant velocity of 59.67 m/s for the next 0.934 min; (c) Constant negative acceleration of - 11.1 m/s 2 for 3.71 s. What was the total displacement

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  1. 18 January, 16:04
    0
    Total displacement Is 1410.9945 meters

    Explanation:

    The total displacement is as follows.

    The first part = triangle A

    3.9 m/s² for 15.3 sec.

    3.9 * 15.3 = 59.67 m/s

    Second part = rectangle

    15.67 m/s for (0.934*60) sec = 56.04 sec

    Third part = triangle B

    11.1m/s² for 3.71 sec

    11.1 * 3.71 = 41.171 m/s

    The area if these shapes gives the displacement.

    For A = 1/2 (15.3*59.67)

    = 456.4755 m

    For B = 15.67 * 56.04

    = 878.1468 m

    For C = 1/2 (41.171*3.71)

    = 76.372

    Total displacement = 456.4755 + 878.1468 + 76.3722

    = 1410.9945 meters
  2. 18 January, 17:37
    0
    a) 456.48m

    b) 3343.91m

    c) - 76.376m

    Total displacement = 3724.01m

    Explanation:

    Using kinematic equation for displacement

    ◇x1 = Vo t + 1/2 at^2

    Where ◇x1 = displacement

    Vo = initial velocity

    a = acceleration

    t = time

    ◇x1 = 0 (15.3) + 1/2 (3.9) (15.3) ^2

    ◇x1 = 0 + 456.48

    Displacement = 456.48m

    b) ◇x2 = Vt1

    V = velocity = 59.67m/s

    ◇x2 = 59.67 * 0.934minute*60seconds

    ◇x2 = 59.67 * 56.04 = 3343.91

    c)

    a = - 11.1m/s

    = (-11.1 m/s^2 * 3.71seconds)

    ◇ x3 = 1/2 (41.18m/s * 3.71)

    ◇x3 = - 76.376m

    Total displacement = ◇x1 + ◇x2 + ◇x3

    Total displacement = 456.48 + 3343.91 - 76.376

    Total displacement = 3724.01m
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