# 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

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