Antiderivative for sin ⁡ ( t 2 ) / 2?I'm taking the AP Calculus BC Exam next week and ran into this problem with no idea how to solve it. Unfortunately, the answer key didn't provide explanations, and I'd really, really appreciate it if someone could explain how to solve this problem.The acceleration of a particle moving along the line is given by a ( 4 ) = t cos ⁡ ( t 2 ). If at time t = 0, its velocity is 2m and position is 4m, what is the position of the particle at time t = 7?(Calculator question; the correct answer is 18.303m .I got that the equation for the velocity ( v = sin ⁡ ( t 2 ) / 2 + 2 m) but I can't seem to find its antiderivative and get the equation for the position.

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Antiderivative for   sin  ⁡  (      t    2    )      /    2?I&#039;m taking the AP Calculus BC Exam next week and ran into this problem with no idea how to solve it. Unfortunately, the answer key didn&#039;t provide explanations, and I&#039;d really, really appreciate it if someone could explain how to solve this problem.The acceleration of a particle moving along the line is given by   a  (  4  )  =  t  cos  ⁡  (      t    2    ). If at time   t  =  0, its velocity is 2m and position is 4m, what is the position of the particle at time   t  =  7?(Calculator question; the correct answer is 18.303m .I got that the equation for the velocity (  v  =  sin  ⁡  (      t    2    )      /    2  +  2  m) but I can&#039;t seem to find its antiderivative and get the equation for the position.

sniokd · Accepted Answer

Step 1Fresnel integral S(z) is defined through the following integral representation      S    (  z  )  =      ∫    0    z    sin  ⁡      (          π      2              t      2        )          d    tand             S        ′    (  z  )  =  sin  ⁡      (          π      2              z      2        )  .Step 2So you have for   v  (  t  )  =      1    2    sin  ⁡  (      t    2    )  +      v    0  .  x  (  t  )  −      x    0    =      ∫    0    t    v  (  τ  )      d    τ  =      1    2        ∫    0    t    sin  ⁡  (      τ    2    )        d    τ  +      v    0    tthat is  x  (  t  )  =      1    2              π      2            S        (                  2        π                  t    )    +      v    0    t  +      x    0  and at   t  =  7  x  (  7  )  =      x    0    +  7      v    0    +      1    2              π      2                                                S                                (                                          2                π                                                  7            )                          ⏟                    ≈      0.4835      where we&#039;ve used WolframAlpha.

Antiderivative for sin ⁡ ( t 2 </msup> ) <mrow cla

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