The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an instant when the magnitude of t

Question

The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an instant when the magnitude of the total acceleration is 6.0 m/s2, what is the speed of the particle? Group of answer choices

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Thu Cúc 3 years 2021-08-13T18:08:39+00:00 1 Answers 239 views 0

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  1. Maris
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    2021-08-13T18:09:54+00:00
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    Answer:

    The speed of the particle is 2.86 m/s

    Explanation:

    Given;

    radius of the circular path, r = 2.0 m

    tangential acceleration,  a_t = 4.4 m/s²

    total magnitude of the acceleration, a = 6.0 m/s²

    Total acceleration is the vector sum of  tangential acceleration and radial acceleration

    a = \sqrt{a_c^2 + a_t^2}\\\\

    where;

    a_c is the radial acceleration

    a = \sqrt{a_c^2 + a_t^2}\\\\a^2 = a_c^2 + a_t^2\\\\a_c^2 = a^2 -a_t^2\\\\a_c = \sqrt{a^2 -a_t^2}\\\\a_c = \sqrt{6.0^2 -4.4^2}\\\\a_c = \sqrt{16.64}\\\\a_c = 4.08 \ m/s^2

    The radial acceleration relates to speed of particle in the following equations;

    a_c = \frac{v^2}{r}

    where;

    v is the speed of the particle

    v^2 = a_c r\\\\v= \sqrt{a_c r} \\\\v = \sqrt{4.08 *2}\\\\v = 2.86 \ m/s

    Therefore, the speed of the particle is 2.86 m/s

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