# Kinematics Using Integral Calculus

• Jan 14th 2011, 08:57 AM
quantoembryo
Kinematics Using Integral Calculus
The problem is stated as follows: The acceleration of a rocket travelling upward is given by $a=(6+0.02s) m/s$, where s is in meters. Determine the time needed for the rocket to reach an altitude of s=100m. Initially, v=0, s=0, t=0.

$vdv=ads$

$\int{_{v_0} ^v}vdv=\int{_{s_0} ^s}(6+0.02s)ds$

$\frac{1}{2}v^2=6s+0.01s^2$

$v=\sqrt{12s+0.02s^2}$

Then, we have:
$dt=ds/v$

$\int dt=\int ds/v$

$t=\int ds[12s+0.02s^2]^{-1}$

and it is at this point my mind draws a blank.... I am taking this course at the same time as integral calculus math is being tought to me, and so I really have not much to go off of for this. Any suggestions?
• Jan 14th 2011, 09:05 AM
Ackbeet
As you can see here, it looks like completing the square in the denominator, followed by a trig substitution, will conquer your integral. Click Show Steps to see how WolframAlpha does it.
• Jan 14th 2011, 09:14 AM
quantoembryo
Yikes, I'm going to assume that there is an easier way to execute this question than that.. This is supposed to be basic integration-based kinematics questions.
• Jan 14th 2011, 09:21 AM
Ackbeet
Completing the square is a high-school-level algebra topic, and trig substitutions are standard techniques taught in freshman integral Calculus.
• Jan 14th 2011, 09:25 AM
quantoembryo
Yes, but we have not even got to substition yet in integration.. We are currently just learning about the definition of the integral. And we most certainly have not seen anything regarding integrating hyperbolic functions
• Jan 14th 2011, 09:52 AM
Ackbeet
Hmm. Well, here's another way to solve your problem: view it as a standard, second-order differential equation (which is what it is, of course). But that's probably even more advanced than the trig substitution, etc. Take a look at this wiki page for trig substitutions, and take a look here around Post # 12 for substitutions in general.