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15 February, 00:19

The velocity function (in meters per second) is given for a particle moving along a line. v (t) = 5t - 9, 0 ≤ t ≤ 3 (a) Find the displacement. - 9/2 Correct: Your answer is correct. m (b) Find the distance traveled by the particle during the given time interval.

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  1. 15 February, 00:48
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    a) The displacement is - 4.5 m.

    b) The traveled distance is 11.7 m.

    Explanation:

    Hi there!

    a) The velocity of the particle is the derivative of the displacement function, x (t):

    v (t) = dx/dt = 5t - 9

    Separating varibles:

    dx = (5t - 9) dt

    Integrating both sides from x = x0 to x and from t = 0 to t.

    x - x0 = 1/2 · 5t² - 9t

    x = 1/2 · 5t² - 9t + x0

    If we place the origin of the system of reference at x = x0, the displacement at t = 3 will be x (3):

    x (3) = 1/2 · 5 · (3) ² - 9 (3) + 0

    x (3) = - 4.5

    The displacement at t = 3 s is - 4.5 m. It means that the particle is located 4.5 m to the left from the origin of the system of reference at t = 3 s.

    b) When the velocity is negative, the particle moves to the left. Let's find the time at which the velocity is less than zero:

    v = 5t - 9

    0 > 5t - 9

    9/5 > t

    1.8 s > t

    Then until t = 1.8 s, the particle moves to the left from the origin of the reference system.

    Let's find the position of the particle at that time:

    x = 1/2 · 5t² - 9t

    x = 1/2 · 5 (1.8 s) ² - 9 (1.8 s)

    x = - 8.1 m

    From t = 0 to t = 1.8 s the traveled distance is 8.1 m. After 1.8 s, the particle has positive velocity. It means that the particle is moving to the right, towards the origin. If at t = 3 the position of the particle is - 4.5 m, then the traveled distance from x = - 8.1 m to x = - 4.5 m is (8.1 m - 4.5 m) 3.6 m.

    Then, the total traveled distance is (8.1 m + 3.6 m) 11.7 m.
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