# The acceleration of a rocket traveling upward 2

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:

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:

$a ds= vdv$

Take the integral of both sides:

$\,\displaystyle\int_{s_0}^{s}a ds= \int_{v_0}^{v}vdv$

$\,\displaystyle\int_{0}^{s} (6 + 0.02s)=\int_{0}^{v}v dv$

(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:

$v=\dfrac{ds}{dt}$

$dt=\dfrac{ds}{v}$

Again, take the integral of both sides:

$\,\displaystyle\int_{0}^{t} dt=\int_{0}^{s}\dfrac{ds}{v}$

(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

$v=322.5$ m/s

$t=19.27$ s

## 2 thoughts on “The acceleration of a rocket traveling upward”

• Zeynep Uçar

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?

• questionsolutions Post author

If integrals give you trouble, try using symbolab as a tool to figure out all the steps 🙂