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14 January, 08:27

Consider the position of a particle moving in three dimensional space to be given by the position vector r = 2.5 t 3 i - 10 t 2 j + 4.00 to, where r is meters and t in seconds. What is the magnitude of the acceleration of the particle at t = 1.0 s?

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  1. 14 January, 08:49
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    25m/s²

    Explanation:

    Given the position of a particle by the three dimensional vectors;

    r = 2.5 t³ i - 10 t² j + 4.00t k where r is in meters and r in seconds, we will get the magnitude of the velocity (v) first since velocity is rate of change in displacement.

    v = dr/dt where r is the position or displacement of an object

    Velocity is gotten by differentiating the given function with respect to time to have;

    v = 7.5t² i - 20t j + 4 k

    Acceleration (a) is the rate of change in velocity.

    a = dv/dt

    Acceleration is derived by differentiating the velocity function with respect to time to have;

    a = 15t i - 20j

    Acceleration of the particle at t = 1.0s will be;

    a = 15 (1) i - 20j

    a = 15i - 20j

    Magnitude of the acceleration can be gotten by finding the resultant of the resulting vector

    a = √ (15i) ² + (20j) ²

    Since i. i = j. j = 1

    a = √15²+20²

    a = √225+400

    a = √625

    a = 25m/s²

    Therefore the magnitude of the acceleration of the particle at t = 1.0 s is 25m/s²
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