Integration Word Problem - Time for raindrop to fall

Vertciel

Junior Member
Joined
May 13, 2007
Messages
78
Hello there,

I am having trouble understanding the solution to this word problem (http://answers.yahoo.com/question/?qid= ... 930AAtAJpe). Unfortunately, the poster wrote his/her solution very messily so I would appreciate a simplified explanation.

Thank you!

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73. Since raindrops grow as they fall, their surface area increases and therefore the resistance to their falling increases. A raindrop has an initial downward velocity of 10 m/s and its downward acceleration is:

\(\displaystyle a=\left\{\begin{array}{cc}9 - 0.9t,&\mbox{ if }0\leq t\leq 10\\0, & \mbox{ if } t>10\end{array}\right.\)

If the raindrop is initially 500 m above the ground, how long does it take to fall?

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My Work (what I understand):

I am allowing the downward direction to be positive.

I know that the antiderivative of acceleration is velocity, so:

\(\displaystyle v=\left\{\begin{array}{cc}9t - 0.45t^2 + C_0,&\mbox{ if }0\leq t\leq 10\\C_1, & \mbox{ if } t>10\end{array}\right.\)

Since it is given that the initial downward velocity is 10 m/s, \(\displaystyle v(0) = 10 m/s = C_0\).

Therefore:

\(\displaystyle v=\left\{\begin{array}{cc}9t - 0.45t^2 + 10,&\mbox{ if }0\leq t\leq 10\\C_1, & \mbox{ if } t>10\end{array}\right.\)

However, I do not understand why \(\displaystyle v(10) = C_1\), etc.
 
Vertciel said:
Hello there,

I am having trouble understanding the solution to this word problem (http://answers.yahoo.com/question/?qid= ... 930AAtAJpe). Unfortunately, the poster wrote his/her solution very messily so I would appreciate a simplified explanation.

Thank you!

---

73. Since raindrops grow as they fall, their surface area increases and therefore the resistance to their falling increases. A raindrop has an initial downward velocity of 10 m/s and its downward acceleration is:

\(\displaystyle a=\left\{\begin{array}{cc}9 - 0.9t,&\mbox{ if }0\leq t\leq 10\\0, & \mbox{ if } t>10\end{array}\right.\)

If the raindrop is initially 500 m above the ground, how long does it take to fall?

--

My Work (what I understand):

I am allowing the downward direction to be positive.

I know that the antiderivative of acceleration is velocity, so:

\(\displaystyle v=\left\{\begin{array}{cc}9t - 0.45t^2 + C_0,&\mbox{ if }0\leq t\leq 10\\C_1, & \mbox{ if } t>10\end{array}\right.\)

Since it is given that the initial downward velocity is 10 m/s, \(\displaystyle v(0) = 10 m/s = C_0\).

Therefore:

\(\displaystyle v=\left\{\begin{array}{cc}9t - 0.45t^2 + 10,&\mbox{ if }0\leq t\leq 10\\C_1, & \mbox{ if } t>10\end{array}\right.\)

However, I do not understand why \(\displaystyle v(10) = C_1\),

Do you understand that the velocity will not change after 10 sec time ?

If you accept that then - at 10 sec whatever the velocity was ( from v(t) ? 10) that will be maintained for rest of the flight.


etc.
 
Thanks for your response.

Will velocity not change after 10 seconds because there will be 0 acceleration after 10 seconds?
 
Vertciel said:
Thanks for your response.

Will velocity not change after 10 seconds because there will be 0 acceleration after 10 seconds? <<< Correct

Now do you understand why do you have [sub:3utujbf1]\(\displaystyle \text v(10) \, = C_1\)[/sub:3utujbf1] ?
 
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