velocity (1 Viewer)

stressedadfff · Sep 23, 2021

hey guys can someone please explain the following part d

Screen Shot 2021-09-23 at 10.58.51 pm.png

specificagent1 · Sep 23, 2021

distance is equal to the total area under the velocity curve, you may not be able to run it as one integral from 0 to 3 as there is postive and negative areas

stressedadfff · Sep 23, 2021

specificagent1 said:
distance is equal to the total area under the velocity curve, you may not be able to run it as one integral from 0 to 3 as there is postive and negative areas

oh how would you suggest then

specificagent1 · Sep 23, 2021

stressedadfff said:
oh how would you suggest then

so the first derivative (velocity) is 2 - 3/(t+1) right?

and the graph looks like this:

Esse

Screen Shot 2021-09-23 at 11.06.15 pm.png

ntially, we want the area under the curve from 0 to 3 so we would run the first | integral from 0 to 0.5 | (absoluted because of negative areas) + integral from 0.5 to 3

CM_Tutor · Sep 24, 2021

The faster way is to consider the position of the particle at the start and finish of the period of interest, plus at any times in the interval when it stops (hence part (c)) as it may change directions.

$\begin{align*} \text{At}\ t = 0\ \text{s},\ x &= 2t - 3\log_e{(t+1)} \\ &= 2(0) - 3\log_e{(0+1)} \\ &= 0 - 3\log_e{1} \\ &= 0\ \text{m} \qquad \text{as $\log_e{1}=0$} \end{align*}$

$\begin{align*} \text{At}\ t = 3\ \text{s},\ x &= 2t - 3\log_e{(t+1)} \\ &= 2(3) - 3\log_e{(3+1)} \\ &= 6 - 3\log_e{4} \\ &= 6 - 6\log_e{2}\ \text{m} \qquad \text{as $\log_e{4} = \log_e{\left(2^2\right)}=2\log_e{2}$} \end{align*}$

From part (c), we know that the particle is stationary at $t=\cfrac{1}{2}\ \text{s}$ , so check

$\begin{align*} \text{At}\ t = \cfrac{1}{2}\ \text{s},\ x &= 2t - 3\log_e{(t+1)} \\ &= 2\left(\cfrac{1}{2}\right) - 3\log_e{\left(\cfrac{1}{2}+1\right)} \\ &= 1 - 3\log_e{\left(\cfrac{3}{2}\right)} \\ &= 1 - 3\log_e{3} + 3\log_e{2}\ \text{m} \qquad \text{as $\log_e{\left(\cfrac{3}{2}\right)} = \log_e{3} - \log_e{2}$} \end{align*}$

So, the particle's travel path:

$\text{starts (at $t = 0$ s) at $x = 0$ m}$
$\text{then moves to $x = 1 - 3\log_e{3} + 3\log_e{2} \approx -0.216...$ m by $t = \cfrac{1}{2}$ s}$
$\text{and then proceeds to $x = 6 - 6\log_e{2} \approx 1.841...$ m by $t = 3$ s}$

And so, over this period of 3 s,

$\begin{align*} \text{the distance travelled by the particle is}\ &= \bigg[0 - \big(1 - 3\log_e{3} + 3\log_e{2}\big)\bigg] + \bigg[6 - 6\log_e{2} - \big(1 - 3\log_e{3} + 3\log_e{2}\big)\bigg] \\ &= -1 + 3\log_e{3} - 3\log_e{2} + 6 - 6\log_e{2} - 1 + 3\log_e{3} - 3\log_e{2} \\ &= \big(-1 + 6 - 1\big) + \big(3 + 3\big)\log_e{3} + \big(-3 - 6 - 3\big)\log_e{2} \\ &= 4 + 6\log_e{3} - 12\log_e{2}\ \text{m} \\ &\approx 2.2739...\ \text{m} \end{align*}$

while its total displacement is $6 - 6\log_e{2} \approx 1.841...\ \text{m}$ .

Using the integration approach, you should find:

$\text{Distance}\ = \left|\int_0^{\frac{1}{2}} 2 - \cfrac{3}{t+1}\ dt\right| + \int_{\frac{1}{2}}^3 2 - \cfrac{3}{t+1}\ dt = 4 + 6\log_e{3} - 12\log_e{2}\ \text{m}$

$\text{Displacement}\ = \int_0^3 2 - \cfrac{3}{t+1}\ dt = 6 - 6\log_e{2}\ \text{m}$

velocity (1 Viewer)

stressedadfff

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stressedadfff

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