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 \(\displaystyle v(10) = C_1\), etc.
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 \(\displaystyle v(10) = C_1\), etc.