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.

#### 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 🙂