# 0.2 Motion in one dimension  (Page 15/16)

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$a=\frac{\Delta v}{t}$

where $\Delta v$ is the change in velocity, i.e. $\Delta v={v}_{f}$ - ${v}_{i}$ . Thus we have

$\begin{array}{ccc}\hfill a& =& \frac{{v}_{f}-{v}_{i}}{t}\hfill \\ \hfill {v}_{f}& =& {v}_{i}+at\hfill \end{array}$

We have seen that displacement can be calculated from the area under a velocity vs. time graph. For uniformly accelerated motion the most complicated velocity vs. time graph we can have is a straight line. Look at the graph below - it represents an object with a starting velocity of ${v}_{i}$ , accelerating to a final velocity ${v}_{f}$ over a total time $t$ .

To calculate the final displacement we must calculate the area under the graph - this is just the area of the rectangle added to the area of the triangle. This portion of the graph has been shaded for clarity.

$\begin{array}{ccc}\hfill {\mathrm{Area}}_{▵}& =& \frac{1}{2}b×h\hfill \\ & =& \frac{1}{2}t×\left({v}_{f}-{v}_{i}\right)\hfill \\ & =& \frac{1}{2}{v}_{f}t-\frac{1}{2}{v}_{i}t\hfill \end{array}$
$\begin{array}{ccc}\hfill {\mathrm{Area}}_{\square }& =& \ell ×b\hfill \\ & =& t×{v}_{i}\hfill \\ & =& {v}_{i}t\hfill \end{array}$
$\begin{array}{ccc}\hfill \mathrm{Displacement}& =& {\mathrm{Area}}_{\square }+{\mathrm{Area}}_{▵}\hfill \\ \hfill \Delta x& =& {v}_{i}t+\frac{1}{2}{v}_{f}t-\frac{1}{2}{v}_{i}t\hfill \\ \hfill \Delta x& =& \frac{\left({v}_{i}+{v}_{f}\right)}{2}t\hfill \end{array}$

This equation is simply derived by eliminating the final velocity ${v}_{f}$ in [link] . Remembering from [link] that

${v}_{f}={v}_{i}+at$

$\begin{array}{ccc}\hfill \Delta x& =& \frac{{v}_{i}+{v}_{i}+at}{2}t\hfill \\ & =& \frac{2{v}_{i}t+a{t}^{2}}{2}\hfill \\ \hfill \Delta x& =& {v}_{i}t+\frac{1}{2}a{t}^{2}\hfill \end{array}$

This equation is just derived by eliminating the time variable in the above equation. From [link] we know

$t=\frac{{v}_{f}-{v}_{i}}{a}$

$\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}\left(\frac{{v}_{f}-{v}_{i}}{a}\right)+\frac{1}{2}a{\left(\frac{{v}_{f}-{v}_{i}}{a}\right)}^{2}\hfill \\ & =& \frac{{v}_{i}{v}_{f}}{a}-\frac{{v}_{i}^{2}}{a}+\frac{1}{2}a\left(\frac{{v}_{f}^{2}-2{v}_{i}{v}_{f}+{v}_{i}^{2}}{{a}^{2}}\right)\hfill \\ & =& \frac{{v}_{i}{v}_{f}}{a}-\frac{{v}_{i}^{2}}{a}+\frac{{v}_{f}^{2}}{2a}-\frac{{v}_{i}{v}_{f}}{a}+\frac{{v}_{i}^{2}}{2a}\hfill \\ \hfill 2a\Delta x& =& -2{v}_{i}^{2}+{v}_{f}^{2}+{v}_{i}^{2}\hfill \\ \hfill {v}_{f}^{2}& =& {v}_{i}^{2}+2a\Delta x\hfill \end{array}$

This gives us the final velocity in terms of the initial velocity, acceleration and displacement and is independent of the time variable.

A racing car is travelling north. It accelerates uniformly covering a distance of 725 m in 10 s. If it has an initial velocity of 10 m $·$ s ${}^{-1}$ , find its acceleration.

1. We are given:

$\begin{array}{ccc}\hfill {v}_{i}& =& 10\phantom{\rule{4pt}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\hfill \\ \hfill \Delta x& =& 725\phantom{\rule{4pt}{0ex}}\mathrm{m}\hfill \\ \hfill t& =& 10\phantom{\rule{4pt}{0ex}}\mathrm{s}\hfill \\ \hfill a& =& ?\hfill \end{array}$
2. If you struggle to find the correct equation, find the quantity that is not given and then look for an equation that has this quantity in it.

$\Delta x={v}_{i}t+\frac{1}{2}a{t}^{2}$
3. $\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}t+\frac{1}{2}a{t}^{2}\hfill \\ \hfill 725\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}& =& \left(10\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}×10\phantom{\rule{3.33333pt}{0ex}}\mathrm{s}\right)+\frac{1}{2}\mathrm{a}×{\left(10\phantom{\rule{3.33333pt}{0ex}}\mathrm{s}\right)}^{2}\hfill \\ \hfill 725\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}-100\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}& =& \left(50{\phantom{\rule{3.33333pt}{0ex}}\mathrm{s}}^{2}\right)\phantom{\rule{3.33333pt}{0ex}}\mathrm{a}\hfill \\ \hfill a& =& 12,5\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}·{\mathrm{s}}^{-2}\hfill \end{array}$
4. The racing car is accelerating at 12,5 m $·$ s ${}^{-2}$ north.

A motorcycle, travelling east, starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64 m in 4 s. Calculate

1. its acceleration
2. its final velocity
3. at what time the motorcycle had covered half the total distance
4. what distance the motorcycle had covered in half the total time.
1. We are given:

$\begin{array}{ccc}\hfill {v}_{i}& =& 0\mathrm{m}·{\mathrm{s}}^{-1}\phantom{\rule{1.em}{0ex}}\left(\mathrm{because}\phantom{\rule{2pt}{0ex}}\mathrm{the}\phantom{\rule{2pt}{0ex}}\mathrm{object}\phantom{\rule{2pt}{0ex}}\mathrm{starts}\phantom{\rule{2pt}{0ex}}\mathrm{from}\phantom{\rule{2pt}{0ex}}\mathrm{rest.}\right)\hfill \\ \hfill \Delta x& =& 64\mathrm{m}\hfill \\ \hfill t& =& 4\mathrm{s}\hfill \\ \hfill a& =& ?\hfill \\ \hfill {v}_{f}& =& ?\hfill \\ \hfill t& =& ?\phantom{\rule{1.em}{0ex}}\mathrm{at}\phantom{\rule{2pt}{0ex}}\mathrm{half}\phantom{\rule{2pt}{0ex}}\mathrm{the}\phantom{\rule{2pt}{0ex}}\mathrm{distance}\Delta \mathrm{x}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}32\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}.\hfill \\ \hfill \Delta x& =& ?\phantom{\rule{1.em}{0ex}}\mathrm{at}\phantom{\rule{2pt}{0ex}}\mathrm{half}\phantom{\rule{2pt}{0ex}}\mathrm{the}\phantom{\rule{2pt}{0ex}}\mathrm{time}\phantom{\rule{2pt}{0ex}}\mathrm{t}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}2\phantom{\rule{3.33333pt}{0ex}}\mathrm{s}.\hfill \end{array}$

All quantities are in SI units.

$\Delta x={v}_{i}t+\frac{1}{2}a{t}^{2}$
3. $\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}t+\frac{1}{2}a{t}^{2}\hfill \\ \hfill 64\phantom{\rule{4pt}{0ex}}\mathrm{m}& =& \left(0m·{s}^{-1}×4\phantom{\rule{4pt}{0ex}}\mathrm{s}\right)+\frac{1}{2}a×{\left(4\phantom{\rule{4pt}{0ex}}\mathrm{s}\right)}^{2}\hfill \\ \hfill 64\phantom{\rule{4pt}{0ex}}\mathrm{m}& =& \left(8\phantom{\rule{4pt}{0ex}}{\mathrm{s}}^{2}\right)\mathrm{a}\hfill \\ \hfill a& =& 8\phantom{\rule{2pt}{0ex}}m·{s}^{-2}\phantom{\rule{3.33333pt}{0ex}}\mathrm{east}\hfill \end{array}$
4. We can use [link] - remember we now also know the acceleration of the object.

${v}_{f}={v}_{i}+at$
5. $\begin{array}{ccc}\hfill {v}_{f}& =& {v}_{i}+at\hfill \\ \hfill {v}_{f}& =& 0\mathrm{m}·{\mathrm{s}}^{-1}+\left(8\mathrm{m}·{\mathrm{s}}^{-2}\right)\left(4\phantom{\rule{4pt}{0ex}}\mathrm{s}\right)\hfill \\ & =& 32\phantom{\rule{2pt}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\phantom{\rule{3.33333pt}{0ex}}\mathrm{east}\hfill \end{array}$
6. We can use [link] :

$\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}+\frac{1}{2}a{t}^{2}\hfill \\ \hfill 32\phantom{\rule{4pt}{0ex}}\mathrm{m}& =& \left(0\mathrm{m}·{\mathrm{s}}^{-1}\right)t+\frac{1}{2}\left(8\mathrm{m}·{\mathrm{s}}^{-2}\right){\left(t\right)}^{2}\hfill \\ \hfill 32\phantom{\rule{4pt}{0ex}}\mathrm{m}& =& 0+\left(4\mathrm{m}·{\mathrm{s}}^{-2}\right){t}^{2}\hfill \\ \hfill 8\phantom{\rule{4pt}{0ex}}{\mathrm{s}}^{2}& =& {t}^{2}\hfill \\ \hfill t& =& 2,83\phantom{\rule{3.33333pt}{0ex}}\mathrm{s}\hfill \end{array}$
7. Half the time is 2 s, thus we have ${v}_{i}$ , $a$ and $t$ - all in the correct units. We can use [link] to get the distance:

$\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}t+\frac{1}{2}a{t}^{2}\hfill \\ & =& \left(0\right)\left(2\right)+\frac{1}{2}\left(8\right){\left(2\right)}^{2}\hfill \\ & =& 16\phantom{\rule{2pt}{0ex}}\mathrm{m}\phantom{\rule{2pt}{0ex}}\mathrm{east}\hfill \end{array}$
1. The acceleration is $8\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-2}$ east
2. The velocity is $32\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-1}$ east
3. The time at half the distance is $2,83\phantom{\rule{2pt}{0ex}}\mathrm{s}$
4. The distance at half the time is $16\phantom{\rule{2pt}{0ex}}\mathrm{m}$ east

where we get a research paper on Nano chemistry....?
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There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
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Preparation and Applications of Nanomaterial for Drug Delivery
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I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
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Nasa has use it in the 60's, copper as water purification in the moon travel.
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nanocopper obvius
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industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
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if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
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analytical skills graphene is prepared to kill any type viruses .
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Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
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write examples of Nano molecule?
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The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
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Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
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What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
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Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
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biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
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