# Thread: Applications Of Logarithms

1. ## Applications Of Logarithms

The displacement, s (m), of a particle P moving in a straight line, at time t seconds, from a fixed point O, is given by: s= -2.3+1.5log(t+2), for t greater than or equal to 0. Find the distance travelled in the first 40 seconds.

Any help would be appreciated!!
Thank you.

2. ## Re: Applications Of Logarithms

The velocity is \displaystyle \begin{align*} \frac{\mathrm{d}s}{\mathrm{d}t} = \frac{3}{2\left( t + 2 \right) } \end{align*}, so the distance travelled is

\displaystyle \begin{align*} d &= \int_0^{40}{ \frac{\mathrm{d}s}{\mathrm{d}t}\,\mathrm{d}t } \\ &= \int_0^{40}{ \frac{3}{2\left( t + 2 \right) } \,\mathrm{d}t } \\ &= \frac{3}{2}\, \left[ \ln{ \left( t + 2 \right) } \right] _0^{40} \\ &= \frac{3}{2}\, \left[ \ln{\left( 42 \right) } - \ln{ \left( 2 \right) } \right] \\ &= \frac{3}{2} \, \ln{ \left( \frac{42}{2} \right) } \\ &= \frac{3}{2} \, \ln{ \left( 21 \right) } \textrm{ m} \end{align*}

3. ## Re: Applications Of Logarithms

Originally Posted by APK911
The displacement, s (m), of a particle P moving in a straight line, at time t seconds, from a fixed point O, is given by: s= -2.3+1.5log(t+2), for t greater than or equal to 0. Find the distance travelled in the first 40 seconds.
Looking at the graph of $s(t)=-2.3+1.5\log(t+2)$, the function is strictly increasing from $t=0$ to $t=40$.

In this case, distance traveled would be $d=s(40)-s(0)$

4. ## Re: Applications Of Logarithms

Thank you very much!!!!

5. ## Re: Applications Of Logarithms

In other words, exactly the same answer that I gave...

6. ## Re: Applications Of Logarithms

Originally Posted by Prove It
In other words, exactly the same answer that I gave...
Correct ... just making the point that the integral $\displaystyle \int_0^{40} \dfrac{3}{2(t+2)} \, dt$ is displacement, which in this case happens to be equal to distance traveled since $v(t) > 0$ in the time interval given.

If an object's velocity changed sign within a given time interval (indicating a change in direction of motion), then distance traveled would be $\displaystyle d = \int_{t_1}^{t_2} |v(t)| \, dt$