Blog: 1) A car starts from rest and moves along the X axis with constant acceleration `5 m/s^2` for `8` seconds. If it then continues will constant...

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When the acceleration $\displaystyle{a}$ is constant, the speed $\displaystyle{V}$ is equal to $\displaystyle{V}{\left({t}\right)}={V}_{{0}}+{a}{t},$ where $\displaystyle{V}_{{0}}$ is the initial speed. The distance traveled is $\displaystyle{D}{\left({t}\right)}={V}_{{0}}{t}+\frac{{{a}{{t}}^{{2}}}}{{2}}.$

1) The final speed is $\displaystyle{V}_{{1}}={V}_{{0}}{t}+{a}{t}={0}+{5}\cdot{8}={40}{\left(\frac{{m}}{{s}}\right)}.$ The distance of accelerated motion is $\displaystyle{0}+\frac{{{5}\cdot{{8}}^{{2}}}}{{2}}={160}{\left({m}\right)},$ the distance of constant speed motion is $\displaystyle{V}_{{1}}\cdot{\left({12}-{8}\right)}={40}\cdot{4}={160}{\left({m}\right)}.$ So the total distance is 320m.

4) (The same formulas): probably at both periods the acceleration is constant and is the same.

For the...

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4) (The same formulas): probably at both periods the acceleration is constant and is the same.

For the first period $\displaystyle{D}_{{1}}={20}={0}+\frac{{{a}\cdot{{2}}^{{2}}}}{{2}},$ so $\displaystyle{a}={10}{\left(\frac{{m}}{{{s}}^{{2}}}\right)},$ and $\displaystyle{V}_{{1}}={0}+{a}\cdot{2}={20}{\left(\frac{{m}}{{s}}\right)}.$

The data for the second period give us nothing new, although there is no contradiction: $\displaystyle{D}_{{2}}={160}={V}_{{1}}\cdot{4}+\frac{{{10}\cdot{{4}}^{{2}}}}{{2}}={80}+{80}$ is true.

So after $\displaystyle{7}$ seconds the speed will be $\displaystyle{a}\cdot{7}={70}{\left(\frac{{m}}{{s}}\right)}.$

2) This is a different but simple story. The average speed is by definition (the total distance traveled) / (total time spent). In our case it is

$\displaystyle\frac{{{2}{A}{B}}}{{\frac{{{A}{B}}}{{30}}+\frac{{{A}{B}}}{{20}}}}=\frac{{2}}{{\frac{{1}}{{30}}+\frac{{1}}{{20}}}}=\frac{{{2}\cdot{20}\cdot{30}}}{{{20}+{30}}}={24}{\left(\frac{{m}}{{s}}\right)}.$