# acceleration/ velocity word problem?

• Feb 16th 2010, 10:40 PM
yoman360
acceleration/ velocity word problem?
A particle moves along the x-axis so that at any time t > 0, its acceleration is given by a(t) = $ln(1+2^t)$. If the velocity of the particle is 2 at time t = 1, then what is the velocity of the particle at time t = 2

I know what to do but i need help.
I know I have to take the integral of a(t) = $ln(1+2^t)$

But i dont know how to evaluate:

$\int ln(1+2^t) dt$

integral ln(1+2^x) - Wolfram|Alpha
wolf fram alpha found the integral but it doesnt do me any good because it comes out all funky
• Feb 17th 2010, 08:12 AM
skeeter
Quote:

Originally Posted by yoman360
A particle moves along the x-axis so that at any time t > 0, its acceleration is given by a(t) = $ln(1+2^t)$. If the velocity of the particle is 2 at time t = 1, then what is the velocity of the particle at time t = 2

I know what to do but i need help.
I know I have to take the integral of a(t) = $ln(1+2^t)$

But i dont know how to evaluate:

$\int ln(1+2^t) dt$

integral ln(1+2^x) - Wolfram|Alpha
wolf fram alpha found the integral but it doesnt do me any good because it comes out all funky

$
v(2) = v(1) + \int_1^2 a(t) \, dt
$

• Feb 18th 2010, 08:02 AM
chisigma
The 'hard part' of the work is of course the computation of the integral...

$\int_{1}^{2} \ln (1+2^{t})\cdot dt$ (1)

With the substitution of variable $x=1+2^{t} \rightarrow dt= \frac{dx} {\ln 2\cdot (x-1)}$ the integral becomes...

$\frac{1}{\ln 2}\cdot \int_{3}^{5} \frac{\ln x}{x-1}\cdot dx$ (2)

... and with further substitution $\xi= \frac{1}{x} \rightarrow dx= - \frac{ d\xi}{\xi^{2}}$ the integral becomes...

$- \frac{1}{\ln 2}\cdot \int_{\frac{1}{5}}^{\frac{1}{3}} \frac{\ln \xi}{\xi \cdot (1- \xi)}\cdot d\xi$ (3)

To be continued in next post...

Kind regards

$\chi$ $\sigma$
• Feb 18th 2010, 08:51 AM
chisigma
In the previous post we are arrived to the identity...

$\int_{1}^{2} \ln (1+2^{t})\cdot dt = -\frac{1}{\ln 2}\cdot \int_{\frac{1}{5}}^{\frac{1}{3}} \frac{\ln \xi}{\xi\cdot (1-\xi)}\cdot d\xi$ (1)

Now for $0<\xi<1$ is...

$\frac{\ln \xi}{\xi\cdot (1-\xi)} = \ln \xi \cdot (\frac {1}{\xi} + 1 + \xi + \xi^{2} + \dots )$ (2)

... and we can apply the following integration rules...

$\int \frac{\ln \xi}{\xi}\cdot d\xi= \frac{\ln^{2} \xi}{2} + c$

$\int \xi^{n}\cdot \ln \xi\cdot d\xi= \xi^{n+1}\{\frac{\ln \xi}{n+1}- \frac{1}{(n+1)^{2}}\} + c$ (3)

... and the well known series expansion...

$\sum_{n=1}^{\infty}\frac{\xi^{n}}{n} = -\ln (1-\xi)$ (4)

... to arrive at the result...

$\int \frac{\ln \xi}{\xi\cdot (1-\xi)}\cdot dx = \frac{\ln^{2} \xi}{2}- \ln \xi \cdot \ln (1-\xi) - \ln \xi \cdot \sum_{n=1}^{\infty} \frac{\xi^{n}}{n^{2}} + c$ (4)

The series in (4) for $\xi= \frac{1}{3}$ and $\xi=\frac{1}{5}$ has very fast convergence so that using (4) the numerical computation of integral (1) is not to hard (Shake) ...

Kind regards

$\chi$ $\sigma$