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.

The acceleration of a rocket traveling upward

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:

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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

 

Final Answers:

v=322.5 m/s

t=19.27 s

 

This question can be found in Engineering Mechanics: Dynamics (SI edition), 13th edition, chapter 12, question 12-13.

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