A particle is moving
with acceleration a of t equals 24t
plus 8 meters per second squared.
The position at time t is s of 0 equals 9 meters
and the velocity at time t equals 0
is v of 0 equals 8 meters per second.
Determine the particles position at t equals 15 seconds.
Notice here we are given the acceleration function
and we need to recover the velocity function,
and then recover the position function
to answer the question.
Remember if s of t is a position function,
then s prime of t is equal to the velocity function
and s double prime of t equals the prime of t
which equals the acceleration function.
So because we are given the acceleration function,
if we find the antiderivative
of the acceleration function,
we can recover the velocity function,
because again the acceleration function
is equal to the derivative of the velocity function.
And when we determine an antiderivative
or evaluate an indef integral,
we are undoing the derivative process.
We'll begin by determining the indef integral
of the acceleration function,
which is the same as evaluating the indef integral
of the derivative of the velocity function.
Since again, the derivative of the velocity function
is acceleration function, which we know
is equal to the indef integral of 24t plus 8.
And by evaluating this indef integral,
we can recover the velocity function.
Well, the antiderivative is equal to 24 times t square
divided by 2 plus 8t plus C.
Simplifying, we have 12t squared plus 8t plus C.
So now we know the velocity function, v of t,
must be in the form of 12t squared plus 8t plus C.
And now looking back at the given information
because we know v of 0 equals 8,
we can determine the specific velocity function.
If v of 0 is equal to 8,
if we substitute 0 for t,
the velocity function must equal 8.
This gives us the equation 12 times the square of 0
plus 8 times 0 plus C must equal 8.
Notice on left side simplifies to just C,
we have C equals 8.
Which means the specific velocity function
with the given conditions is v of t
equals 12t squared plus 8t plus 8.
Let's go ahead and record this above.
And now from here,
if we find the antiderivative of the velocity function,
we can recover the position function
because the velocity function
is equal to the derivative of the position function.
So let's do that next.
Again, evaluating the indef integral
of the velocity function
is the same as evaluating the indef integral
of the derivative of the position function,
which is equal to the indef integral
of the quantity 12t squared plus 8t plus 8 dt.
And again, because the indef integral
undoes a derivative process,
this will help us determine the position function
so that we can then answer the question.
So the antiderivative is equal to 12 times t cubed
divided by 3 plus 8 times t squared
divided by 2 plus 8t plus C.
Simplifying, we now know
the form of the position function, s of t,
must be 4t cubed
plus 4t squared
plus 8t plus C.
But going back to the given information,
we know s of 0 equals 9.
Knowing s of 0 equals 9
allows us to determine the value of C,
which will give us the exact position function
so that we can then determine the position
at t equals 15 seconds.
So if s of 0 equals 9,
we have the equation 4 times the cube of 0
plus 4 times the square of 0
plus 8 times 0 plus C must equal 9.
Again, on the left, everything simplifies to 0 except C
giving us C equals 9,
now we know the position function is s of t
equals 4t cubed
plus 4t squared plus 8t plus 9.
Let's also record this above.
And now that we finally have the position function,
we can determine the position at t equals 15 seconds.
s of 15 is equal to 4 times the cube of 15
plus 4 times the square of 15
plus 8 times 15 plus 9.
The particles position at 15 seconds is 14529 meters.
I hope you found this helpful.