Two antiderivative problems!!!

**Kayla_N** · April 6th 2009, 04:42 PM

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?

**Reckoner** · April 6th 2009, 05:01 PM

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).

**Kayla_N** · April 6th 2009, 08:23 PM

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

**Reckoner** · April 6th 2009, 08:32 PM

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?

**Kayla_N** · April 6th 2009, 08:53 PM

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!!!

**Reckoner** · April 6th 2009, 09:02 PM

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.

**Kayla_N** · April 6th 2009, 09:10 PM

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

**Reckoner** · April 6th 2009, 09:14 PM

Originally Posted by Kayla_N

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

Where did you get 739 from?

**Kayla_N** · April 6th 2009, 09:17 PM

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.

**Reckoner** · April 6th 2009, 09:21 PM

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 $t=3?$

**Kayla_N** · April 6th 2009, 09:24 PM

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

**Reckoner** · April 6th 2009, 09:29 PM

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 $s(3).$

**Kayla_N** · April 6th 2009, 09:35 PM

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.

Thanks for your help...

**Reckoner** · April 6th 2009, 09:40 PM

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!

**Kayla_N** · April 6th 2009, 09:42 PM

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

Thread: Two antiderivative problems!!!

LinkBack

Thread Tools

Display

Two antiderivative problems!!!

Similar Math Help Forum Discussions

antiderivative problems

Please help me with antiderivative/l'hopitale problems

Antiderivative problems!!

Antiderivative Word Problems

A few antiderivative application problems

Search Tags