An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s2 as it goes through. The station is 210 m long.
(a) How long is the nose of the train in the station?
(b) How fast is it going when the nose leaves the station?
(c) If the train is 130 m long, when does the end of the train leave the station?
(d) What is the velocity of the end of the train as it leaves?
Solution:
Part A
We are given the following: v_0=22.0\:\text{m/s}; a=-0.150\:\text{m/s}^2; and \Delta x=210\:\text{m}
We are required to solve for time, t. We are going to use the formula
\Delta x=v_0t+\frac{1}{2}at^2 Substituting the given values, we have
\begin{align*}
\Delta x & =v_0t+\frac{1}{2}at^2 \\
210\:\text{m} & =\left(22.0\:\text{m/s}\right)t+\frac{1}{2}\left(-0.150\:\text{m/s}^2\right)t^2
\end{align*}If we simplify and rearrange the terms into a general quadratic equation, we have
Solve for t using the quadratic formula. We are given a=0.075;\:b=-22t;\:c=210.
\begin{align*}
t & =\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\
t& =\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\left(0.075\right)\left(210\right)}}{2\left(0.075\right)}\\
\end{align*}There are two values of t that can satisfy the quadratic equation.
t=9.88\ \text{s} \qquad \text{and} \qquad t=283.46 \ \text{s}
Discard t=283.46 \ \text{s} as it can be seen from the problem that this is not a feasible solution. So, we have
t=9.88\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)Part B
We have the same given values from part a. We are going to solve v_f in the formula
\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x
So we have
\begin{align*}
\left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\
v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\
v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(210\:\text{m}\right)} \\
v_f & =20.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part C
We are given the same values as in the previous parts, except that for the value of \Delta x since we should incorporate the length of the train. For the distance, \Delta x, we have
\begin{align*}
\Delta x & =210\:\text{m}+130\:\text{m} \\
\Delta x & = 340 \ \text{m}
\end{align*}To solve for time t, we are going to use the formula
\Delta x=v_ot+\frac{1}{2}at^2If we rearrange the formula into a general quadratic equation and solve for t using the quadratic formula, we come up with
t=\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a}Substituting the given values:
\begin{align*}
t & =\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a} \\
t & =\frac{-\left(22.0\:\text{m/s}\right)\pm \sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)}}{-0.150\:\text{m/s}^2} \\
t &=16.4\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part D
We have the same given values. We are going to solve for v_f in the equation
\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x
Substituting the given values:
\begin{align*}
\left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\
v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\
v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)} \\
v_f & =19.5\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
You must be logged in to post a comment.