Chapter 2, Problem 65GP

(a)

To determine

To Find:The time taken to make a trip of $9.0\text{ km}$ in $6$ stations.

Answer to Problem 65GP

  $T=8.80\text{ min}$

Explanation of Solution

Given:

Acceleration of the train, $a=1.1{\text{ m/s}}^2$

Speed of the train, $v=90\text{ km/h = 25 m/s}$

Deacceleration of the train, $a_d=-2.0{\text{ m/s}}^2$

Stoppage time at each station, $t_s=20\text{ s}$

Initial speed of train, $u=0$

Distance between the stations, $d=1.8\text{ km = 1800 m}$

Formula used:

Third equation of motion:

  $v^2=u^2+2as$

First equation of motion:

  $v=u+at$

Where,

  $v=$ Final speed.

  $u=$ Initial speed.

  $a=$ Acceleration.

  $s=$ Distance covered.

Calculation:

There are total 6 stations including initial and final, therefore, there are 4 intermediate stations at which train stops.

Total stoppage time:

  $t=4t_s=4\times 20=80\text{ s}$

Distance travelled by train during acceleration can be calculated by third equation of motion:

  $\begin{array}{c}{S}_a=\frac{v^2-u^2}{2a}\\ =\frac{{\left(25\right)}^2-0^2}{2\times 1.1}\\ =284.1\text{ m}\end{array}$

Time taken by train during acceleration:

  $\begin{array}{c}{t}_a=\frac{v-u}{a}\\ =\frac{25-0}{1.1}\\ =22.73\text{ s}\end{array}$

Distance travelled by train during deacceleration can be calculated by third equation of motion:

  $\begin{array}{c}{S}_d=\frac{v^2-u^2}{2a_d}\\ =\frac{{\left(25\right)}^2-0^2}{2\times \left(-2.0\right)}\\ =-156.25\text{ m}\\ =\text{156}\text{.25 m}\end{array}$

Time taken by train during deacceleration:

  $\begin{array}{c}{t}_d=\frac{v-u}{a_d}\\ =\frac{25-0}{2.0}\\ =12.5\text{ s}\end{array}$

Distance travelled by train during constant speed:

  $\begin{array}{c}{S}_c=d-\left(S_a+S_d\right)\\ =1800-\left(284.1+156.25\right)\\ =1359.65\text{ m}\end{array}$

Time taken by train during constant speed:

  $\begin{array}{c}{t}_c=\frac{1359.65}{25}\\ =54.386\text{ s}\end{array}$

Total time between two stations:

  $\begin{array}{c}{t}_{total}=t_a+t_d+t_c\\ =22.73+12.5+54.386\\ =89.62\text{ s}\end{array}$

Total trips of the train are 5, therefore total time of journey:

  $\begin{array}{c}T=5t_{total}+t\\ =5\times 89.62+80\\ =528.1\text{ s}\\ =528.1\times \frac{1\text{ min}}{60\text{ s}}\text{s}\\ T=8.80\text{ min}\end{array}$

Conclusion:

Hence, the time taken to make a trip of $9.0\text{ km}$ in $6$ stations: $\text{8}\text{.80 min}$

(b)

To determine

To Find: The time taken to make a trip of $9.0\text{ km}$ in $4$ stations.

Answer to Problem 65GP

  $T=7.55\text{ min}$

Explanation of Solution

Given:

Acceleration of the train, $a=1.1{\text{ m/s}}^2$

Speed of the train, $v=90\text{ km/h = 25 m/s}$

Deacceleration of the train, $a_d=-2.0{\text{ m/s}}^2$

Stoppage time at each station, $t_s=20\text{ s}$

Initial speed of train, $u=0$

Distance between the stations, $d=3\text{ km = 3000 m}$

Formula used:

Third equation of motion:

  $v^2=u^2+2as$

First equation of motion:

  $v=u+at$

Where,

  $v=$ Final speed.

  $u=$ Initial speed.

  $a=$ Acceleration.

  $s=$ Distance covered.

Calculation:

There are total 4 stations including initial and final, therefore, there are 2 intermediate stations at which train stops.

Total stoppage time:

  $t=2t_s=2\times 20=40\text{ s}$

Distance travelled by train during acceleration can be calculated by third equation of motion:

  $\begin{array}{c}{S}_a=\frac{v^2-u^2}{2a}\\ =\frac{{\left(25\right)}^2-0^2}{2\times 1.1}\\ =284.1\text{ m}\end{array}$

Time taken by train during acceleration:

  $\begin{array}{c}{t}_a=\frac{v-u}{a}\\ =\frac{25-0}{1.1}\\ =22.73\text{ s}\end{array}$

Distance travelled by train during deacceleration can be calculated by third equation of motion:

  $\begin{array}{c}{S}_d=\frac{v^2-u^2}{2a_d}\\ =\frac{{\left(25\right)}^2-0^2}{2\times \left(-2.0\right)}\\ =-156.25\text{ m}\\ =\text{156}\text{.25 m}\end{array}$

Time taken by train during deacceleration:

  $\begin{array}{c}{t}_d=\frac{v-u}{a_d}\\ =\frac{25-0}{2.0}\\ =12.5\text{ s}\end{array}$

Distance travelled by train during constant speed:

  $\begin{array}{c}{S}_c=d-\left(S_a+S_d\right)\\ =3000-\left(284.1+156.25\right)\\ =2559.65\text{ m}\end{array}$

Time taken by train during constant speed:

  $\begin{array}{c}{t}_c=\frac{2559.65}{25}\\ =102.386\text{ s}\end{array}$

Total time between two stations:

  $\begin{array}{c}{t}_{total}=t_a+t_d+t_c\\ =22.73+12.5+102.386\\ =137.62\text{ s}\end{array}$

Total trips of the train are 3, therefore total time of journey:

  $\begin{array}{c}T=3t_{total}+t\\ =3\times 137.62+40\\ =452.86\text{ s}\\ =452.86\times \frac{1\text{ min}}{60\text{ s}}\text{s}\\ =7.55\text{ min}\end{array}$

Conclusion:

Hence, the time taken to make a trip of $9.0\text{ km}$ in $4$ stations: $7.55\text{ min}$