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

Question

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

1 Answer

Impact of this question

3279 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-19T21:34:38

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

Explanation:

Displacement is given by:

$s (t) = t - ln (2 t + 1)$ $s(t)=t-ln(2t+1)$

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

$s (2) - s (0) = 2 ln 5$ $s(2) - s(0) = 2 ln 5$ .

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

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

$t - ln (2 t + 1) = 0$ $t-ln(2t+1) = 0$

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

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

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

So, just to make it clear:

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

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

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

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

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

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

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

So the distance $s_{0, 2}$ $s_(0,2)$ moved during the period $t \in [0, 2]$ $t in [0,2]$ is:

$s_{0, 2} = 2 - ln 5 + ∣ ∣ ∣ \frac{1}{2} - ln 2 ∣ ∣ ∣$ $s_(0,2) = 2 - ln 5 + abs(1/2 - ln 2)$

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

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

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