# Problems with integration word problems

• Apr 25th 2010, 04:31 PM
bobsanchez
Problems with integration word problems
I was absent the day that we covered these, and have yet to have received a sufficient explanation of how to do them. I understand integration fairly well, but I'm unsure of how to exactly apply it in these problems. One excellent example is this:

The Washington Monument is around 550 ft. tall. If you're trying to throw an object from gound level to the top of the Monument, what would your initial velocity need to be? Gravity is at -32 ft/s, don't worry about air resistance.

I understand how to integrate, but I don't know how to begin to apply it to this kind of a problem. Thanks in advance for all the advice.
• Apr 25th 2010, 04:58 PM
skeeter
Quote:

Originally Posted by bobsanchez
I was absent the day that we covered these, and have yet to have received a sufficient explanation of how to do them. I understand integration fairly well, but I'm unsure of how to exactly apply it in these problems. One excellent example is this:

The Washington Monument is around 550 ft. tall. If you're trying to throw an object from gound level to the top of the Monument, what would your initial velocity need to be? Gravity is at -32 ft/s, don't worry about air resistance.

I understand how to integrate, but I don't know how to begin to apply it to this kind of a problem. Thanks in advance for all the advice.

$\displaystyle a = -32$

antiderivative of acceleration is velocity ...

$\displaystyle v = -32t + C$

at $\displaystyle t = 0$ , velocity = $\displaystyle v_0$ , initial velocity.

$\displaystyle v = v_0 - 32t$

antiderivative of velocity is position ...

$\displaystyle y = v_0 t - 16t^2 + C$

if the starting position is the ground, $\displaystyle C = 0$

$\displaystyle y = v_0 t - 16t^2$

height of the monument is 550 ft ...

$\displaystyle 550 = v_0 t - 16t^2$

at the top of its trajectory, $\displaystyle v = 0$ ... $\displaystyle t = \frac{v_0}{32}$

solve for $\displaystyle v_0$
• Apr 25th 2010, 05:11 PM
bobsanchez
So you basically just integrate acceleration, then subtract that from what your velocity, obtain the position, and solve for whatever it is that you're looking for? Okay, that makes sense.
• Apr 25th 2010, 05:30 PM
skeeter
Quote:

Originally Posted by bobsanchez
So you basically just integrate acceleration, then subtract that from what your velocity, obtain the position, and solve for whatever it is that you're looking for? Okay, that makes sense.

huh?
• Apr 25th 2010, 05:34 PM
bobsanchez
Quote:

Originally Posted by skeeter
huh?

Oops. Initial velocity, rather.
• Apr 25th 2010, 05:39 PM
skeeter
Quote:

Originally Posted by bobsanchez
Oops. Initial velocity, rather.

no "subtraction" ... the negative sign is there to indicate the direction of the acceleration (down).

understand that $\displaystyle v_0$ is the constant of integration for the antiderivative of $\displaystyle a = -32$.