The acceleration of a rocket traveling upward is given by a = (6 + 0.02s) m/s^2, where s is in meters. Determine the rocket’s velocity when s = 2 km and the time needed to reach this altitude. Initially, v = 0 and s = 0 when t = 0.
Image from: R. C. Hibbeler, K. B. Yap, and S. C. Fan, Mechanics for Engineers: Dynamics (SI Edition), 13th ed. Singapore: Pearson Education South Asia, 2013.
Solution:
Show me the final answer↓
As the acceleration is not constant, we will need to integrate our acceleration to figure out the velocity. To do so, we will need to use the following equation:
Take the integral of both sides:
(For the acceleration integral, the lower limit is 0 because the rocket starts at a height of 0 m. For the velocity integral, remember that the rocket starts from rest, meaning the lower limit is 0 m/s.)
(6s+\dfrac{0.02s^2}{2})\Big|_0^{s}=\dfrac{v^2}{2}\Big|_0^v
6s+\dfrac{0.02s^2}{2}=\dfrac{v^2}{2}
v=\sqrt{12s+0.02s^2}
When the height is 2000 m, the velocity is:
v=\sqrt{12(2000)+0.02(2000)^2}
v=322.5 m/s
To find the time, remember that:
dt=\dfrac{ds}{v}
Again, take the integral of both sides:
(substitute the velocity equation we found)
\,\displaystyle\int_{0}^{t} dt=\int_{0}^{2000}\dfrac{ds}{\sqrt{12s+0.02s^2}}if it’s hard to visualize the right side of this integral, remember that you can write it like so:
\,\displaystyle\int_{0}^{t} dt=\int_{0}^{2000}\dfrac{1}{\sqrt{12s+0.02s^2}}ds
(This is a complicated integral, however, you can see the integral solved here: https://goo.gl/iWgq1f)
t=19.27 s
Final Answers:
t=19.27 s
Have a nice day. I’m trying to understand the solution of the integral in the last step, but I can’t. Can you help with this?
If integrals give you trouble, try using symbolab as a tool to figure out all the steps 🙂