A car starts from a certain point at a velocity of 100 kph. After 20 seconds, another car starting from rest, starts at the same point following...

Thursday, February 21, 2008

A car starts from a certain point at a velocity of 100 kph. After 20 seconds, another car starting from rest, starts at the same point following...

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Certainly car B will overtake car B. After some time the speed of car B will become greater than the speed of car A and a quicker object always overtakes a slower one.

Now determine the specific time. The distance travelled by car A is

$\displaystyle{d}_{{A}}={V}_{{A}}\cdot{t},$ where $\displaystyle{V}_{{A}}={100}\frac{{{k}{m}}}{{h}}=\frac{{{100},{000}{m}}}{{{3600}{s}}}\approx{27.8}\frac{{m}}{{s}}$ and time $\displaystyle{t}$ is measured from the start.

The distance travelled by car B is

$\displaystyle{d}_{{B}}={a}_{{B}}\cdot\frac{{{\left({t}-{20}\right)}}^{{2}}}{{2}},$ where $\displaystyle{a}_{{B}}={3}\frac{{m}}{{{s}}^{{2}}}.$

We have to find $\displaystyle{t}_{{1}}\gt{20}$ such that $\displaystyle{d}_{{A}}{\left({t}_{{1}}\right)}={d}_{{B}}{\left({t}_{{1}}\right)}:$

$\displaystyle{27.8}\cdot{t}={3}\frac{{{\left({t}-{20}\right)}}^{{2}}}{{2}},$ or

$\displaystyle{3}{{t}}^{{2}}-{120}{t}-{2}\cdot{27.8}{t}+{1200}={0},$ or $\displaystyle{3}{{t}}^{{2}}-{175.6}{t}+{1200}={0}.$

So $\displaystyle{t}_{{1}}=\frac{{{175.6}+\sqrt{{{{\left({175.6}\right)}}^{{2}}-{4}\cdot{3}\cdot{1200}}}}}{{6}}\approx$ 50.6 (s). The solution with minus before the root is less than 20.

The distance both travel will be $\displaystyle{d}_{{A}}={27.8}\cdot{50.6}\approx$ 1407 (m).

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Thursday, February 21, 2008

A car starts from a certain point at a velocity of 100 kph. After 20 seconds, another car starting from rest, starts at the same point following...

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