Question

when brakes are applied the speed of a train decreases from 96 kmh to 48 kmh in 800m. how much further wil the train moves before coming to rest?

Answer 1

.

Task.

When brakes are applied the speed of a train decreases from $v_{0}=96~{}km/h$ to $v_{1}=48~{}km/h$ in $d_{0}=800~{}m$. How much further will the train moves before coming to rest?

Solution.

Suppose that the train moves with constant deceleration $a<0$. If the brakes were applied at time $t=0$, then at time $t$ the train will pass the distance

$d(t)=v_{0}t+\frac{at^{2}}{2}$

and the velocity of the train at that time is

$v(t)=v_{0}+at.$

Thus we have a system of the following two equations:

\[ \begin{cases}d_{0}=v_{0}t+\frac{at^{2}}{2}\\

v_{1}=v_{0}+at\end{cases} \]

Let us first find $t$ and $a$ for which $d(t)=d_{0}$ and $v(t)=v_{1}$.

From the second equation we get:

$at=v_{1}-v_{0}.$

Substituting it into the first equation we obtain

$d_{0}=v_{0}t+\frac{at\cdot t}{2}=v_{0}t+\frac{(v_{1}-v_{0})t}{2}=\frac{2v_{0}t+v_{1}t-v_{0}t}{2}=\frac{(v_{0}+v_{1})t}{2}.$

Hence

$t=\frac{2d_{0}}{v_{0}+v_{1}}.$

Substituting this expression for $t$ into the second equation we obtain the formula for $a$:

$a=\frac{v_{1}-v_{0}}{t}=\frac{(v_{1}-v_{0})(v_{0}+v_{1})}{2d_{0}}=\frac{v_{1}^{2}-v_{0}^{2}}{2d_{0}}.$

Now we will find the time $\bar{t}$, when the train will stop. It can be derived from the relation

$v(\bar{t})=0.$

From the second equation we have

$0=v(\bar{t})=v_{0}+a\bar{t}$

whence

$\bar{t}=-\frac{v_{0}}{a}=-\frac{2d_{0}v_{0}}{v_{1}^{2}-v_{0}^{2}}=\frac{2d_{0}v_{0}}{v_{0}^{2}-v_{1}^{2}}.$

Hence the total distance passed by the train after applying brakes is

$d(\bar{t})=v_{0}\bar{t}+\frac{a\bar{t}^{2}}{2}.$

Taking to account that

$a\bar{t}=-v_{0},$

and using the formula for $\bar{t}$, we obtain that

$d(\bar{t})=v_{0}\bar{t}+\frac{a\bar{t}\cdot\bar{t}}{2}=v_{0}\bar{t}-\frac{v_{0}\bar{t}}{2}=\frac{v_{0}\bar{t}}{2}=\frac{d_{0}v_{0}^{2}}{v_{0}^{2}-v_{1}^{2}}.$

Therefore after passing $d_{0}$ the train will move to the rest the distance

$d_{1}=d(\bar{t})-d_{0}=\frac{d_{0}v_{0}^{2}}{v_{0}^{2}-v_{1}^{2}}-d_{0}=\frac{d_{0}v_{0}^{2}}{v_{0}^{2}-v_{1}^{2}}-\frac{d_{0}(v_{0}^{2}-v_{1}^{2})}{v_{0}^{2}-v_{1}^{2}}=\frac{d_{0}v_{1}^{2}}{v_{0}^{2}-v_{1}^{2}}.$

Substituting values we obtain:

$d_{1}=\frac{d_{0}v_{1}^{2}}{v_{0}^{2}-v_{1}^{2}}=\frac{0.8\cdot 48^{2}}{96^{2}-48^{2}}\approx 0.267~{}km=267~{}m.$

Answer.

267 $m$