The acceleration vector of a particle Q is given by the following acceleration vector a = 4i – 6tJ + sin(.2t)k m/s^2, where t is the time. A

Question

The acceleration vector of a particle Q is given by the following acceleration vector a = 4i – 6tJ + sin(.2t)k m/s^2, where t is the time. At t=0, the pointe Q is located at (x0,y0,z0) = (1,3,-5)m, and has a velocity vo= 2i -7j +8.4k m/s. Find the speed of Q and its distance from the starting point at t= 3s.

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Khải Quang 6 months 2021-07-27T16:18:01+00:00 1 Answers 3 views 0

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  1. thnahdat
    0
    2021-07-27T16:19:45+00:00
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    Answer:

    V(3) =14i^ -34j^ +8.57 k^

    S(3) =(25,-45,3.97)

    Explanation:

    We know that

    V =a dt

    from t=0 to 3s

    V = 4i – 6tJ + sin(.2t)k m/s² dt

    V =4t i^ – 3t^2j^ – cos(2t)/2 k^ +C

    So we have

    V(0) =-1/2 k^ +C =2i -7j +8.4 k

    C=2i -7j +8.9k^

    V =4t i^ – 3t^2j^ – cos(2t)/2 k^ + 2i -7j +8.9k^

    Then putting 3s

    V(3) =14i^ -34j^ +8.57 k^

    Also

    S(t) =V(t) dt

    S(t) = 4t i^ – 3t^2j^ – cos(2t)/2 k^ + 2i -7j +8.9k^] dt

    S(t)= (2t^2 +2t)i^ – (t^3 +7t) j^ -[sin(2t)/4 – 8.9t] k^ +C =i+3j-5k

    So at t= 0 we have

    S(0) = C= i+3j-5k

    So S(t) =(2t^2 +2t)i^ – (t^3 +7t) j^ -[sin(2t)/4 – 8.9t] k^ +i+3j-5k

    When t= 3

    S(3) =25i -45j + 3.97k

    S(3) =(25,-45,3.97)

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