$F=Ma$ Olympiad Problem 1

Let the length of a train car be $l$ . We know that the displacement $s$ of the train travelling with constant acceleration $a$ passes the observer after time $t$ is given by:

$s = ut + \frac{1}{2}at^2 \quad \Rightarrow s = \frac{1}{2}at^2$

where $u=0$ is the initial velocity of the train.

Since it takes $5$ s for the first car to pass by, we have:

$l = \frac{1}{2}a(5^2) \quad \Rightarrow a = \frac {2}{25}l$

From $s = \frac{1}{2}at^2 = \frac {l}{25}t^2 \quad \Rightarrow t(s) = 5\sqrt{\frac{s}{l}}$ .

So the time for the $10^{th}$ to pass is given by:

$t(10l) - t(9l) = 5\sqrt{\frac{10l}{l}} - 5\sqrt{\frac{9l}{l}} = 5(\sqrt{10}-3) \approx \boxed{0.81}$ s.

Let the length of each car be = $s$

For the first car, $u=0$ and $t=5$

Let acceleration be equal to $a$

Thus, $\quad \quad \quad s\quad =\quad ut\quad +\quad \frac { 1 }{ 2 } a{ t }^{ 2 }\\ \Rightarrow \quad s\quad =\quad (0\times 5)\quad +\quad \frac { 1 }{ 2 } a({ 5 }^{ 2 })\\ \Rightarrow \quad s\quad =\quad \frac { 25a }{ 2 } \\ \Rightarrow \quad \frac { 2s }{ 25 } \quad =\quad a\\$

The velocity $v$ acquired after the passing of the ${9}^{th}$ car would be given by ${ \quad \quad v }^{ 2 }\quad =\quad { u }^{ 2 }\quad +\quad 2as\\ \Rightarrow { v }^{ 2 }\quad =\quad 0\quad +\quad 2\left( \frac { 2s }{ 25 } \right) \left( 9s \right) \\ \Rightarrow { v }^{ 2 }\quad =\quad \quad \frac { 4{ s }\times 9s }{ 25 } \\ \Rightarrow { v }^{ 2 }\quad =\quad \frac { 36{ s }^{ 2 } }{ 25 } \\ \Rightarrow v\quad =\quad \frac { 6s }{ 5 }$

Now, for the time taken by the ${10}^{th}$ car to pass, we use the equation

$s\quad =\quad ut\quad +\quad \frac { 1 }{ 2 } a{ t }^{ 2 }$

Applying the values, we get

$\quad \quad s\quad =\quad \frac { 6s }{ 5 } t\quad +\quad \frac { 1 }{ 2 } \frac { 2s }{ 25 } { t }^{ 2 }\\ \Rightarrow s\quad =\quad \frac { 6st }{ 5 } \quad +\quad \frac { s{ t }^{ 2 } }{ 25 } \\ \Rightarrow s\quad =\quad \frac { 30st\quad +\quad s{ t }^{ 2 } }{ 25 } \\ \Rightarrow 1\quad =\quad \frac { 30t\quad +\quad { t }^{ 2 } }{ 25 } \\ \Rightarrow 25\quad =\quad 30t\quad +\quad t^{ 2 }$

On solving the equation, we get

$t\quad =\quad 5(-3-\sqrt { 10 } )\\ t\quad =\quad 5(\sqrt { 10 } -3)$

Since time cannot be negative, we only have the second value of $t$

Thus $t = 5(\sqrt { 10 } -3) = 5(3.16-3) = 5 \times .16 = \boxed{.80}$

The distance is proportional to the square of the time. So, if the length of one train takes 5 seconds, 9 trains will take 5 sqrt(9) and 10 will take 5 sqrt(10). Therefore the tenth train will take 5*sqrt(10) - 15 = 0.81