a) A light-rail commuter train accelerates at a rate of 1.35 m/s². How long does it take to reach its top speed of 80.0 km/h, starting from rest?

b) The same train ordinarily decelerates at a rate of 1.65 m/s². How long does it take to come to a stop from its top speed?

c) In emergencies, the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s²?

Solution:

Part A

We are given the following: a=1.35 \ \text{m/s}^2; v_f=80.0 \ \text{km/h}; and v_0=0 \ \text{m/s}.

From the formula v_f=v_0+at, we can solve for t as

t=\frac{v_f-v_0}{a}

We need to convert 80.0 km/h to m/s first so that we have a unit uniformity for all the given values.

\begin{align*}
 80\:\text{km/hr} & =\left(80\:\text{km/hr}\right)\left(\frac{1000\:\text{m}}{1\:\text{km}}\right)\left(\frac{1\:\text{hr}}{3600\:\text{s}}\right) \\
& =22.2222\:\text{m/s}
\end{align*}

Substituting the given values into the formula, we have

\begin{align*}
t & =\frac{v_f-v_0}{a}
\\
t & =\frac{22.2222\:\text{m/s}-0\:\text{m/s}}{1.35\:\text{m/s}^2} \\
t & =16.5\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

For this problem, we still use the formula used in Part (a). This time, the values of the final and initial velocities interchange and the value of the given deceleration is negative of acceleration. The given values are a=-1.65 \ \text{m/s}^2; v_0=22.2222 \ \text{m/s}[/katex]; and v_f=0 \ \text{m/s}

\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{0\:\text{m/s}-22.2222\:\text{m/s}}{-1.65\:\text{m/s}^2} \\
t & =13.5\:\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

For this problem, we will use the formula

\begin{align*}
a & =\frac{v_f-v_0}{\Delta t}
\end{align*}

Substituting all the given values into the formula, we have

\begin{align*}
a & =\frac{0\:\text{m/s}-22.2222\:\text{m/s}}{8.30\:\text{s}} \\
a & =2.68\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}