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.