Ask Question
26 July, 03:49

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

+5
Answers (2)
  1. 26 July, 04:49
    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. 26 July, 05:42
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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) ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers