# Thread:

1. ## Two antiderivative problems!!!

1. Evaluate the indefinite integral.

2. A stone is thrown straight up from the edge of a roof, feet above the ground, at a speed of feet per second.

A. Remembering that the acceleration due to gravity is , how high is the stone seconds later?

B. At what time does the stone hit the ground?

C. What is the velocity of the stone when it hits the ground?

2. Originally Posted by Kayla_N
1. Evaluate the indefinite integral.

Let $u=5+14x^4\Rightarrow du=56x^3\,dx.$

2. A stone is thrown straight up from the edge of a roof, feet above the ground, at a speed of feet per second.

A. Remembering that the acceleration due to gravity is , how high is the stone seconds later?
Integrate the acceleration function to get the velocity function $v.$ You can solve for the constant of integration by noting that $v(0)=20.$

Integrate $v$ to get the stone's position function $s.$ Set $s(0)=775$ and you can solve for the constant of integration. Then just evaluate $s(3).$

B. At what time does the stone hit the ground?
That is, for what value of $t$ is $s(t)=0?$

C. What is the velocity of the stone when it hits the ground?
Evaluate the velocity function at the time found in part (B).

3. I still dont understand what you wrote for number 2..can you explain a little bit more.???

4. Originally Posted by Kayla_N
I still dont understand what you wrote for number 2..can you explain a little bit more.???
Velocity is the derivative of position; acceleration is the derivative of velocity.

The stone is undergoing a constant $-32\text{ ft./s}^2$ acceleration due to gravity, so the acceleration function is

$a(t)=-32.$

The velocity function is therefore

$v(t)=\int a(t)\,dt=\int(-32)\,dt=-32t+C_0.$

We know that the initial velocity is 20 ft./s, so

$v(0)=20\Rightarrow -32\cdot0+C_0=20\Rightarrow C_0=20$

and

$v(t)=20-32t.$

Work similarly to find the position function. Can you continue?

5. Originally Posted by Reckoner
Velocity is the derivative of position; acceleration is the derivative of velocity.

The stone is undergoing a constant $-32\text{ ft./s}^2$ acceleration due to gravity, so the acceleration function is

$a(t)=-32.$

The velocity function is therefore

$v(t)=\int a(t)\,dt=\int(-32)\,dt=-32t+C_0.$

We know that the initial velocity is 20 ft./s, so

$v(0)=20\Rightarrow -32\cdot0+C_0=20\Rightarrow C_0=20$

and

$v(t)=20-32t.$

Work similarly to find the position function. Can you continue?

since v(t)=20-32t, so t= .625.
Then to find s(t)= \int (-32+20)dt
s(t)=-16t^2+20t+C2
3=-16(0)+20(0)+c2
C2=3??
Then i dont know what else to do!!!

6. Originally Posted by Kayla_N
since v(t)=20-32t, so t= .625.
At $t=0.625,$ the velocity is zero (i.e., the stone is not moving). This will occur when the stone reaches its maximum height. Were you asked to find this?

Then to find s(t)= \int (-32+20)dt
s(t)=-16t^2+20t+C2
3=-16(0)+20(0)+c2
What are you doing? The initial height is 775 ft., not 3. After you solve for $C_2,$ evaluate $s(3)$ to get the position after 3 seconds.

7. So the s(t) -16t^2+20t+739 right???

8. Originally Posted by Kayla_N
So the s(t) -16t^2+20t+739 right???
Where did you get 739 from?

9. Originally Posted by Reckoner
Where did you get 739 from?
well 739 is wrong, i did again 775=-16t^2+20t+C
then i plug in t=3 and i got C=859?

Jeeze i'm so lost.

10. Originally Posted by Kayla_N
well 739 is wrong, i did again 775=-16t^2+20t+C
then i plug in t=3 and i got C=859?
Think about what you are doing. The stone's initial height (at 0 seconds) is 775 ft., so when $t=0,\;s(t)=775.$ So why are you setting $t=3?$

11. i'm trying to calculate the height at 3 seconds.. Can you please show me how to get part A??

12. Originally Posted by Kayla_N
i'm trying to calculate the height at 3 seconds.. Can you please show me how to get part A??
Do as I said:

We have $s(t)=-16t^2+20t+C_2,$ and $s(0)=775$ (as I mentioned above). Therefore,

$s(0)=-16\cdot0^2+20\cdot0+C_2=775$

$\Rightarrow C_2=775$

and

$s(t)=-16t^2+20t+775.$

Now find $s(3).$

13. ok and i got 691. From there i went on to second part...s(t)=0
0=-16t^2+20t+775. I used quadratic formala and got 7.6127 sec. then moved on to 3 part, i used V= (7.6127)(-32)+20= -223.6068.

14. Originally Posted by Kayla_N
ok and i got 691. From there i went on to second part...s(t)=0
0=-16t^2+20t+775. I used quadratic formala and got 7.6127 sec. then moved on to 3 part, i used V= (7.6127)(-32)+20= -223.6068.
Looks good!

15. Thanks...just part A that i'm stuck on..Thanks again for you help. I really really appreciated.
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