$$a=\frac{\Delta v}{t}$$
where
$\Delta v$ is the change in velocity, i.e.
$\Delta v={v}_{f}$ -
${v}_{i}$ .
Thus we have
$$\begin{array}{ccc}\hfill a& =& \frac{{v}_{f}-{v}_{i}}{t}\hfill \\ \hfill {v}_{f}& =& {v}_{i}+at\hfill \end{array}$$
Derivation of
[link]
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.
$$\begin{array}{ccc}\hfill {\mathrm{Area}}_{\u25b5}& =& \frac{1}{2}b\times h\hfill \\ & =& \frac{1}{2}t\times ({v}_{f}-{v}_{i})\hfill \\ & =& \frac{1}{2}{v}_{f}t-\frac{1}{2}{v}_{i}t\hfill \end{array}$$
$$\begin{array}{ccc}\hfill {\mathrm{Area}}_{\square}& =& \ell \times b\hfill \\ & =& t\times {v}_{i}\hfill \\ & =& {v}_{i}t\hfill \end{array}$$
$$\begin{array}{ccc}\hfill \mathrm{Displacement}& =& {\mathrm{Area}}_{\square}+{\mathrm{Area}}_{\u25b5}\hfill \\ \hfill \Delta x& =& {v}_{i}t+\frac{1}{2}{v}_{f}t-\frac{1}{2}{v}_{i}t\hfill \\ \hfill \Delta x& =& \frac{({v}_{i}+{v}_{f})}{2}t\hfill \end{array}$$
Derivation of
[link]
This equation is simply derived by eliminating the final velocity
${v}_{f}$ in
[link] . Remembering from
[link] that
$${v}_{f}={v}_{i}+at$$
then
[link] becomes
$$\begin{array}{ccc}\hfill \Delta x& =& \frac{{v}_{i}+{v}_{i}+at}{2}t\hfill \\ & =& \frac{2{v}_{i}t+a{t}^{2}}{2}\hfill \\ \hfill \Delta x& =& {v}_{i}t+\frac{1}{2}a{t}^{2}\hfill \end{array}$$
Derivation of
[link]
This equation is just derived by eliminating the time variable in the above equation. From
[link] we know
$$t=\frac{{v}_{f}-{v}_{i}}{a}$$
Substituting this into
[link] gives
$$\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}\left(\frac{{v}_{f}-{v}_{i}}{a}\right)+\frac{1}{2}a{\left(\frac{{v}_{f}-{v}_{i}}{a}\right)}^{2}\hfill \\ & =& \frac{{v}_{i}{v}_{f}}{a}-\frac{{v}_{i}^{2}}{a}+\frac{1}{2}a\left(\frac{{v}_{f}^{2}-2{v}_{i}{v}_{f}+{v}_{i}^{2}}{{a}^{2}}\right)\hfill \\ & =& \frac{{v}_{i}{v}_{f}}{a}-\frac{{v}_{i}^{2}}{a}+\frac{{v}_{f}^{2}}{2a}-\frac{{v}_{i}{v}_{f}}{a}+\frac{{v}_{i}^{2}}{2a}\hfill \\ \hfill 2a\Delta x& =& -2{v}_{i}^{2}+{v}_{f}^{2}+{v}_{i}^{2}\hfill \\ \hfill {v}_{f}^{2}& =& {v}_{i}^{2}+2a\Delta x\hfill \end{array}$$
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
$\xb7$ s
${}^{-1}$ , find its acceleration.
Identify what information is given and what is asked for
We are given:
$$\begin{array}{ccc}\hfill {v}_{i}& =& 10\phantom{\rule{4pt}{0ex}}\mathrm{m}\xb7{\mathrm{s}}^{-1}\hfill \\ \hfill \Delta x& =& 725\phantom{\rule{4pt}{0ex}}\mathrm{m}\hfill \\ \hfill t& =& 10\phantom{\rule{4pt}{0ex}}\mathrm{s}\hfill \\ \hfill a& =& ?\hfill \end{array}$$
Find an equation of motion relating the given information to the acceleration
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.
We can use equation
[link]
$$\Delta x={v}_{i}t+\frac{1}{2}a{t}^{2}$$
Substitute your values in and find the answer
$$\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}t+\frac{1}{2}a{t}^{2}\hfill \\ \hfill 725\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}& =& (10\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\xb7{\mathrm{s}}^{-1}\times 10\phantom{\rule{3.33333pt}{0ex}}\mathrm{s})+\frac{1}{2}\mathrm{a}\times {\left(10\phantom{\rule{3.33333pt}{0ex}}\mathrm{s}\right)}^{2}\hfill \\ \hfill 725\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}-100\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}& =& \left(50{\phantom{\rule{3.33333pt}{0ex}}\mathrm{s}}^{2}\right)\phantom{\rule{3.33333pt}{0ex}}\mathrm{a}\hfill \\ \hfill a& =& 12,5\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}\xb7{\mathrm{s}}^{-2}\hfill \end{array}$$
Quote the final answer
The racing car is accelerating at 12,5 m
$\xb7$ 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
its acceleration
its final velocity
at what time the motorcycle had covered half the total distance
what distance the motorcycle had covered in half the total time.
Identify what information is given and what is asked for
We are given:
$$\begin{array}{ccc}\hfill {v}_{i}& =& 0\mathrm{m}\xb7{\mathrm{s}}^{-1}\phantom{\rule{1.em}{0ex}}\left(\mathrm{because}\phantom{\rule{2pt}{0ex}}\mathrm{the}\phantom{\rule{2pt}{0ex}}\mathrm{object}\phantom{\rule{2pt}{0ex}}\mathrm{starts}\phantom{\rule{2pt}{0ex}}\mathrm{from}\phantom{\rule{2pt}{0ex}}\mathrm{rest.}\right)\hfill \\ \hfill \Delta x& =& 64\mathrm{m}\hfill \\ \hfill t& =& 4\mathrm{s}\hfill \\ \hfill a& =& ?\hfill \\ \hfill {v}_{f}& =& ?\hfill \\ \hfill t& =& ?\phantom{\rule{1.em}{0ex}}\mathrm{at}\phantom{\rule{2pt}{0ex}}\mathrm{half}\phantom{\rule{2pt}{0ex}}\mathrm{the}\phantom{\rule{2pt}{0ex}}\mathrm{distance}\Delta \mathrm{x}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}32\phantom{\rule{3.33333pt}{0ex}}\mathrm{m}.\hfill \\ \hfill \Delta x& =& ?\phantom{\rule{1.em}{0ex}}\mathrm{at}\phantom{\rule{2pt}{0ex}}\mathrm{half}\phantom{\rule{2pt}{0ex}}\mathrm{the}\phantom{\rule{2pt}{0ex}}\mathrm{time}\phantom{\rule{2pt}{0ex}}\mathrm{t}\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}2\phantom{\rule{3.33333pt}{0ex}}\mathrm{s}.\hfill \end{array}$$
All quantities are in SI units.
Acceleration : Find a suitable equation to calculate the acceleration
We can use
[link]
$$\Delta x={v}_{i}t+\frac{1}{2}a{t}^{2}$$
Substitute the values and calculate the acceleration
$$\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}t+\frac{1}{2}a{t}^{2}\hfill \\ \hfill 64\phantom{\rule{4pt}{0ex}}\mathrm{m}& =& (0m\xb7{s}^{-1}\times 4\phantom{\rule{4pt}{0ex}}\mathrm{s})+\frac{1}{2}a\times {\left(4\phantom{\rule{4pt}{0ex}}\mathrm{s}\right)}^{2}\hfill \\ \hfill 64\phantom{\rule{4pt}{0ex}}\mathrm{m}& =& \left(8\phantom{\rule{4pt}{0ex}}{\mathrm{s}}^{2}\right)\mathrm{a}\hfill \\ \hfill a& =& 8\phantom{\rule{2pt}{0ex}}m\xb7{s}^{-2}\phantom{\rule{3.33333pt}{0ex}}\mathrm{east}\hfill \end{array}$$
Final velocity : Find a suitable equation to calculate the final velocity
We can use
[link] - remember we now also know the acceleration of the object.
$${v}_{f}={v}_{i}+at$$
Substitute the values and calculate the final velocity
$$\begin{array}{ccc}\hfill {v}_{f}& =& {v}_{i}+at\hfill \\ \hfill {v}_{f}& =& 0\mathrm{m}\xb7{\mathrm{s}}^{-1}+(8\mathrm{m}\xb7{\mathrm{s}}^{-2})\left(4\phantom{\rule{4pt}{0ex}}\mathrm{s}\right)\hfill \\ & =& 32\phantom{\rule{2pt}{0ex}}\mathrm{m}\xb7{\mathrm{s}}^{-1}\phantom{\rule{3.33333pt}{0ex}}\mathrm{east}\hfill \end{array}$$
Time at half the distance : Find an equation to calculate the time
We can use
[link] :
$$\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}+\frac{1}{2}a{t}^{2}\hfill \\ \hfill 32\phantom{\rule{4pt}{0ex}}\mathrm{m}& =& (0\mathrm{m}\xb7{\mathrm{s}}^{-1})t+\frac{1}{2}(8\mathrm{m}\xb7{\mathrm{s}}^{-2}){\left(t\right)}^{2}\hfill \\ \hfill 32\phantom{\rule{4pt}{0ex}}\mathrm{m}& =& 0+(4\mathrm{m}\xb7{\mathrm{s}}^{-2}){t}^{2}\hfill \\ \hfill 8\phantom{\rule{4pt}{0ex}}{\mathrm{s}}^{2}& =& {t}^{2}\hfill \\ \hfill t& =& 2,83\phantom{\rule{3.33333pt}{0ex}}\mathrm{s}\hfill \end{array}$$
Distance at half the time : Find an equation to relate the distance and time
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:
$$\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}t+\frac{1}{2}a{t}^{2}\hfill \\ & =& \left(0\right)\left(2\right)+\frac{1}{2}\left(8\right){\left(2\right)}^{2}\hfill \\ & =& 16\phantom{\rule{2pt}{0ex}}\mathrm{m}\phantom{\rule{2pt}{0ex}}\mathrm{east}\hfill \end{array}$$
Write down the final results.
The acceleration is
$8\phantom{\rule{2pt}{0ex}}\mathrm{m}\xb7\mathrm{s}{}^{-2}$ east
The velocity is
$32\phantom{\rule{2pt}{0ex}}\mathrm{m}\xb7\mathrm{s}{}^{-1}$ east
The time at half the distance is
$\mathrm{2,83}\phantom{\rule{2pt}{0ex}}\mathrm{s}$
The distance at half the time is
$16\phantom{\rule{2pt}{0ex}}\mathrm{m}$ east