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## Homework Statement

The acceleration of a rocket travelling upward is given by a = (6 + 0.02s) m/(s^2), where s in in meters. Determine the rocket's velocity when s = 2 km = 2000 m and the time needed to reach this altitude. Initially, v = 0 and s = 0 when t = 0.

## Homework Equations

Given: [tex] a = 6 + 0.02s, v_{0} = 0, s_{0} = 0 [/tex]

Find: [tex] v_{|s=2000}, t_{|s=2000} [/tex]

## The Attempt at a Solution

Finding the velocity when the rocket is 2000 m from its starting position didn't give me many problems. I did the following:

[tex] a ds = v dv [/tex]

[tex] (6+0.02s)ds = v dv [/tex]

[tex]\int (6+0.02s)ds = \int v dv[/tex]

Note: I don't know how to do bounded integrals on Latex so the first integral goes from 0 to s, whereas the second integral goes from 0 to v.

[tex]\frac{v^2}{2} = 6s + 0.01s^2[/tex]

Which leads to:

[tex] v = \sqrt{12s+0.02s^2}[/tex]

and [tex] v(2000) = 322.6 m/s [/tex] which is correct.

As for the second part (finding the time t, when s = 2000 m again), this is where I need help:

[tex] v = \frac{ds}{dt} \Rightarrow dt = \frac{ds}{v}. [/tex] Meaning

[tex]\int dt = \int \frac{ds}{v} \Rightarrow [/tex]

[tex]\int dt = \int \frac {ds}{\sqrt{12s+0.02s^2}}[/tex]

(Integral goes from 0 to t, and from 0 to s, respectively.)

This is where I'm stuck. I looked in the Math appendix of my Dynamics book where I got this problem, and according to it, I should have a negative square root when the integral has been evaluated. However, I don't think that should be the case if I did this problem properly.

Any suggestions?

BTW, I'm sorry for the way the formulas are written and my half-use of Latex; I'm sure it could look more readable but it's my first time using Latex so I'm not 100% familiar with it.

Thanks in advance.

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