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Determine properties of a particle with position given by an integral equation

The position of a particle at time t moving along a straight line is given by a function f(t). When 0 \leq t \leq 1 this function is given by

    \[ f(t) = \int_0^t \frac{1 + 2 \sin (\pi x) \cos (\pi x)}{1+x^2} \, dx. \]

When t \geq 1 the particle moves with constant acceleration (the acceleration it obtains at time t = 1 from the above equation of motion). Compute the following. (Do not attempt to evaluate the integral.)

  1. The acceleration of the particle at time t = 2.
  2. The velocity of the particle at time t = 1.
  3. The velocity of the particle when t > 1.
  4. The difference f(t) - f(1) when t > 1.

  1. Since acceleration at time t \geq 1 is constant, the acceleration at time t = 2 is the same as the acceleration at time t = 1. To find the acceleration at time t = 1 we take two derivatives of f(t) and evaluate at t = 1,

        \begin{align*}  && f(t) &= \int_0^t \frac{1 + 2 \sin (\pi x) \cos (\pi x)}{1 + x^2} \, dx \\ \implies && f'(t) &= \frac{1+2 \sin (\pi t) \cos (\pi t)}{1+t^2} \\  &&&= \frac{1 + \sin (2 \pi t)}{1+t^2} \\ \implies && f''(t) &= \frac{(1+t^2)(2 \pi)(\cos (2 \pi t)) - 2t (1 + \sin^2 (\pi t))}{(1+t^2)^2} \\ \implies && f''(1) &= \frac{4 \pi - 2}{4} \\  &&&= \pi - \frac{1}{2}. \end{align*}

  2. From part (a) we already computed f'(t). The velocity at time t = 1 is then

        \[ v(1) = f'(1) = \frac{1}{2}. \]

  3. The velocity at a time t with t > 1 is the velocity at time t = 1 calculated in part (b) plus the amount of time that has passed multiplied by the acceleration over that time. Thus, using parts (a) and (b) we have for t > 1,

        \begin{align*}  v(t) &= f'(1) + \left( \pi - \frac{1}{2} \right) (t-1) \\  &= \frac{1}{2} + \left( \pi - \frac{1}{2} \right) (t-1). \end{align*}

  4. The difference f(t) - f(1) is the position at time t minus the position at time t = 1, where t > 0. So, using the linearity of the integral with respect to the interval of integration we have,

        \[ f(t) - f(1) = \int_1^t f'(x) \, dx, \qquad f'(t) - f'(1) = \int_1^t f''(x) \, dx. \]

    We know f''(t) is the acceleration, and f''(t) for times t > 1 is given by

        \[ f''(t) = \pi - \frac{1}{2}. \]

    Therefore,

        \begin{align*}  &&f'(t) - f'(1) &= \left( \pi - \frac{1}{2} \right) (t-1) \\ \implies && f(t) - f(1) &= \int_1^t (x-1)\left( \pi - \frac{1}{2} \right) \, dx \\  &&&= \int_1^t \left( \pi - \frac{1}{2} \right)x \, dx - \int_1^t \left( \pi - \frac{1}{2} \right) \, dx \\ &&&= \left( \pi - \frac{1}{2} \right)\left( \frac{1}{2} (t^2 -1) \right) - (t-1) \left( \pi - \frac{1}{2} \right) \\ &&&= \left( \pi - \frac{1}{2} \right)(t^2 - 2t + 1)\left( \frac{1}{2} \right) \\ &&&= \left( \pi - \frac{1}{2} \right) \left( \frac{t^2 - 2t + 1}{2} \right). \end{align*}

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