Solution to Question 10
System Setup and Variables
Let us define the coordinate system with the origin at the position of the passenger. The direction of the train’s motion is taken as the positive $x$-axis ($\hat{i}$).
- Let $l$ be the length of a single coach.
- Let $N$ be the total number of coaches in the train.
- Let $a$ be the constant acceleration of the train.
The problem states the passenger is at the beginning of the $n$-th coach ($n=3$) at $t=0$, and the train starts moving from rest ($u=0$). This means the front of Coach 3 is aligned with the passenger at $t=0$.
(a) Calculating the Total Number of Coaches
Since the train starts from rest with constant acceleration $a$, the distance covered $s$ in time $t$ is given by: $$ \vec{s} = \frac{1}{2}\vec{a}t^2 \implies s = \frac{1}{2}at^2 $$
Step 1: Analyze the motion of the 3rd Coach
The 3rd coach ($C_3$) passes the passenger in time $\Delta t_1 = 5.0 \text{ s}$. The length of this segment is $l$. $$ l = \frac{1}{2}a (\Delta t_1)^2 \quad \dots(1) $$
Step 2: Analyze the motion of the rest of the train
The “rest of the train including the 3rd coach” consists of coaches from $n=3$ to the last coach $N$. The number of coaches in this segment is: $$ \text{Count} = N – (n – 1) = N – (3 – 1) = N – 2 $$ The total length of this segment is $(N-2)l$.
Since the train starts from rest at the beginning of this entire interval (when the front of $C_3$ is at the passenger), the time taken to cover this total distance is $\Delta t_2 = 20 \text{ s}$. $$ (N-2)l = \frac{1}{2}a (\Delta t_2)^2 \quad \dots(2) $$
Step 3: Solve for N
Dividing equation (2) by equation (1): $$ \frac{(N-2)l}{l} = \frac{\frac{1}{2}a (\Delta t_2)^2}{\frac{1}{2}a (\Delta t_1)^2} $$ $$ N – 2 = \left( \frac{\Delta t_2}{\Delta t_1} \right)^2 $$
Substituting the given values $\Delta t_1 = 5.0 \text{ s}$ and $\Delta t_2 = 20 \text{ s}$: $$ N – 2 = \left( \frac{20}{5} \right)^2 $$ $$ N – 2 = (4)^2 = 16 $$ $$ N = 18 $$
Therefore, there are 18 coaches in the train.
(b) Time Interval for the Last Coach
We need to find the time interval during which the last coach ($C_{18}$) passes the passenger. Let $\Delta t_{\text{last}}$ be this interval.
This interval is the difference between the time taken for the whole train ($C_3 \dots C_{18}$) to pass and the time taken for the train up to the second-to-last coach ($C_3 \dots C_{17}$) to pass.
- Total distance of $C_3 \dots C_{18}$ is $(18-2)l = 16l$. Time taken is $\Delta t_2 = 20 \text{ s}$.
- Distance of $C_3 \dots C_{17}$ is $(17-2)l = 15l$. Let the time taken be $t’$.
Using the proportionality $t \propto \sqrt{s}$: $$ \frac{t’}{\Delta t_2} = \sqrt{\frac{15l}{16l}} = \frac{\sqrt{15}}{4} $$ $$ t’ = 20 \times \frac{\sqrt{15}}{4} = 5\sqrt{15} \text{ s} $$
Now, calculating the value: $$ t’ \approx 5 \times 3.873 = 19.365 \text{ s} $$
The time interval for the last coach is: $$ \Delta t_{\text{last}} = \Delta t_2 – t’ $$ $$ \Delta t_{\text{last}} = 20 – 19.365 = 0.635 \text{ s} $$
Rounding to two decimal places, the last coach passed in 0.64 s.