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

by RoRi

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*}

3 comments

      • William says:

        I think their mistake on the last step was treating f'(1) as a function instead of a constant. Integral of f'(1) should be tf'(1), not f(1) (Im pretty sure at least – would appreciate if someone else could confirm).

        Though it would probably be easier to do the last part of the problem by finding f(t) by integrating f'(t) and then subtracting f(1) at the end.

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