# If an object has a displacement function s(t)=t-ln(2t+1) where t is in seconds and t>=0, can you find the distance travelled in the first 2 seconds?

Mar 19, 2017

$= \frac{3}{2} + \ln \left(\frac{2}{5}\right)$

#### Explanation:

Displacement is given by:

$s \left(t\right) = t - \ln \left(2 t + 1\right)$

We can see that the change in displacement (the vector quantity that measures spatial differences) during the period $t \in \left[0 , 2\right]$ would simply be:

$s \left(2\right) - s \left(0\right) = 2 \ln 5$.

But that is not necessarily the distance (the scalar quantity that measures spatial differences) travelled by the object in $t \in \left[0 , 2\right]$.

Solving the displacement equation looks tricky, ie we need to solve:

$t - \ln \left(2 t + 1\right) = 0$

I am not sure how you do that simply, so, despite the Pre-Calc classification, from calculus , we can see that velocity is:

$v \left(t\right) = \dot{s} \left(t\right) = 1 - \frac{2}{2 t + 1}$

$v \left(t\right) = 0 \implies 1 - \frac{2}{2 t + 1} = 0 \implies t = \frac{1}{2}$

So, just to make it clear:

$v \left(0\right) = - 1$, object is moving to left

$v \left(\frac{1}{2}\right) = 0$, object has stopped moving

$v \left(2\right) = \frac{3}{5}$, object is moving to right

To obtain the distance travelled, we need to look at the displacements at these 3 points:

$s \left(0\right) = 0$, object is moving to left

$s \left(\frac{1}{2}\right) = \frac{1}{2} - \ln 2 \approx - 0.2$

$s \left(2\right) = 2 - \ln 5 \approx 0.4$, object is moving to right

So the distance ${s}_{0 , 2}$ moved during the period $t \in \left[0 , 2\right]$ is:

${s}_{0 , 2} = 2 - \ln 5 + \left\mid \frac{1}{2} - \ln 2 \right\mid$

$= \frac{3}{2} + \ln \left(\frac{2}{5}\right)$