23 PE - College Physics (Urone)

Question

a) A light-rail commuter train accelerates at a rate of \({\bf{1}}{\bf{.35}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\). 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 \({\bf{1}}{\bf{.65}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\). 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 \({\bf{m/}}{{\bf{s}}^{\bf{2}}}\)?

Step-by-Step Solution

Answer

(a) The duration for acceleration of commuter train is \({\bf{16}}{\bf{.5}}\;{\bf{s}}\).

(b) The duration for acceleration of commuter train is \({\bf{13}}{\bf{.5}}\;{\bf{s}}\).

(c) The emergency deceleration of commuter train is \({\bf{2}}{\bf{.7}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\).

Determination of duration for acceleration of commuter train

(a)

Given Data:

The initial speed of commuter train is \(u = 0\) 

The final speed of commuter train is \(v = 80\;{\rm{km}}/{\rm{h}} = 22.22\;{\rm{m}}/{\rm{s}}\) 

The acceleration of commuter train is \(a = 1.35\;{\rm{m}}/{{\rm{s}}^2}\) 

The deceleration of commuter train is \(f = 1.65\;{\rm{m}}/{{\rm{s}}^2}\) 

The time for emergency braking is \(T = 8.30\;{\rm{s}}\) 

The time for acceleration and deceleration of commuter train is found by using first equation of motion.

The duration for acceleration of commuter train is given as

\(v = u + a{t_a}\) 

Here, \(a\) is the acceleration of commuter train.

Substitute all the values in the above equation.

\(\begin{array}{c}22.22\;{\rm{m}}/{\rm{s}} = 0 + \left( {1.35\;{\rm{m}}/{{\rm{s}}^2}} \right){t_a}\\{t_a} = 16.5\;{\rm{s}}\end{array}\) 

Therefore, the duration for acceleration of commuter train is \(16.5\;{\rm{s}}\).

Determination of deceleration of rocket sled

(b)

The duration for deceleration of commuter train is given as

\(v = u + f \cdot {t_d}\) 

Here, \(f\) is the deceleration of commuter.

Substitute all the values in the above equation.

\(\begin{array}{c}22.22\;{\rm{m}}/{\rm{s}} = 0 + \left( {1.65\;{\rm{m}}/{{\rm{s}}^2}} \right){t_d}\\{t_d} = 13.5\;{\rm{s}}\end{array}\) 

Therefore, the duration for deceleration of commuter train is \(13.5\;{\rm{s}}\).

Determination of emergency deceleration of commuter train

(c)

The deceleration of commuter train is given as

\({f_e} = \frac{v}{T}\) 

Here, \({f_e}\) is the emergency deceleration of commuter train.

Substitute all the values in the above equation.

\(\begin{array}{l}{f_e} = \frac{{22.22\;{\rm{m}}/{\rm{s}}}}{{8.30\;{\rm{s}}}}\\{f_e} = 2.7\;{\rm{m}}/{{\rm{s}}^2}\end{array}\) 

Therefore, the emergency deceleration of commuter train is \(2.7\;{\rm{m}}/{{\rm{s}}^2}\).