An object has a position given by the radius vector r = [2.0 m + (3.00 m/s)t](i)+ [3.0 m – (2.00 m/s^2)t^2](j). Here (i) and (j) are the uni

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An object has a position given by the radius vector r = [2.0 m + (3.00 m/s)t](i)+ [3.0 m – (2.00 m/s^2)t^2](j). Here (i) and (j) are the unit vectors along x and y and all quantities are in SI units. What is the speed and magnitude of the acceleration in m/s^2 of the object at time t = 2.00 s

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Yến Oanh 3 years 2021-08-20T22:18:17+00:00 1 Answers 134 views 0

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  1. vankhanh
    0
    2021-08-20T22:20:05+00:00
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    Answer:

    The speed of the object is (3i - 4.00tj)m/s

    The magnitude of the acceleration is 4.00m/s²

    Explanation:

    Given – position vector;

    r = (2.0 + 3.00t)i + (3.0 – 2.00t²)j       ——————-(i)

    To get the speed vector (v), take the first derivative of equation (i) with respect to time t as follows;

    v = \frac{dr}{dt}

     v = \frac{d[(2.0 + 3.00t)i + (3.0 - 2.00t^2)j] }{dt}  

    v  = 3i - 4.00tj      ————————(ii)

    To get the acceleration vector (a), take the first derivative of the speed vector in equation(ii) as follows;

    a = \frac{dv}{dt}

    a = \frac{d(3i - 4.00tj)}{dt}

    a = -4.00j

    The magnitude of the acceleration |a| is therefore given by

    |a| = |-4.00|

    |a| = 4.00 m/s²

    In conclusion;

    the speed of the object is (3i - 4.00tj)m/s

    the magnitude of the acceleration is 4.00m/s²

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