The position of a particle at time moving along a straight line is given by a function . When this function is given by

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

- The acceleration of the particle at time .
- The velocity of the particle at time .
- The velocity of the particle when .
- The difference when .

- Since acceleration at time is constant, the acceleration at time is the same as the acceleration at time . To find the acceleration at time we take two derivatives of and evaluate at ,
- From part (a) we already computed . The velocity at time is then
- The velocity at a time with is the velocity at time 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 ,
- The difference is the position at time minus the position at time , where . So, using the linearity of the integral with respect to the interval of integration we have,
We know is the acceleration, and for times is given by

Therefore,

In part d you didn’t consider the velocity at time equal to1 second. I think the answer given in the book is correct

You are absolutely right. I guess, the solution creator just forgot the term f'(1) when solving the equation.

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.