A passenger is standing on the platform at the beginning of n-th (= 3rd) coach of a train. If the tr

Q: Why is the distance for the first condition just $L$?

The passenger is standing exactly at the beginning of the $n^{\text{th}}$ coach. For this specific coach to completely pass him, the train must move forward by exactly the length of one coach, which we defined as $L$.

Q: Why is the total distance $(N - n + 1)L$?

The coaches that pass the passenger are the $n^{\text{th}}$, $(n+1)^{\text{th}}$, all the way to the $N^{\text{th}}$ coach. The number of coaches in this sequence is $N - n + 1$. Since each has length $L$, the total distance is $(N - n + 1)L$.

Q: Why can we just divide the two equations?

Dividing the equations is a powerful algebraic trick. Since both equations contain the unknown constants $a$ (acceleration) and $L$ (length), dividing them cancels these unknowns out, leaving an equation with only the variables we know or want to find.

Q: How did we get the time for the last coach?

We found the total time for all coaches to pass ($\Delta t_2$), and the time for all coaches *except* the last one to pass ($t'$). The difference between these two times is the time it took for just the last coach to pass.

Q: Why does the last coach pass so much faster than the third coach?

Because the train is constantly accelerating! By the time the last coach reaches the passenger, the train has been speeding up for nearly 20 seconds, so it's moving much faster than it was when the third coach passed.

Question

A passenger is standing on the platform at the beginning of $n^{	ext{th}}$ ($= 3^{	ext{rd}}$) coach of a train. If the train starts moving with constant acceleration, the third coach passes by the passenger in $\Delta t_1 = 5.0	ext{ s}$ and rest of the train including the $3^{	ext{rd}}$ coach in $\Delta t_2 = 20	ext{ s}$.
(a) How many coaches are in the train?
(b) In what time interval did the last coach pass by the passenger?

Sigma Percentile · Accepted Answer

### The Setup: A Passenger's Perspective

Imagine you are standing on a railway platform. A train is completely stationary in front of you. You happen to be standing exactly at the boundary where the second coach ends and the third coach begins. Suddenly, the train starts moving forward with a constant acceleration.

Because the train starts from rest, its initial velocity $u$ is zero. The distance $S$ it covers in any given time $t$ is governed by the second equation of motion: 
$$S = ut + \frac{1}{2}at^2$$
Since $u=0$, this simplifies beautifully to $S = \frac{1}{2}at^2$. This single equation is the master key to unlocking the entire problem.

### Part A: Counting the Coaches

The problem gives us two distinct observations. First, the third coach (let's call it the $n^{	ext{th}}$ coach, where $n=3$) takes exactly $\Delta t_1 = 5.0	ext{ s}$ to completely pass you. For this to happen, the train must move forward by exactly the length of one coach. Let's denote the length of a single coach as $L$. We can write our first equation:
$$L = \frac{1}{2}a(\Delta t_1)^2$$

Next, we are told that the *entire rest of the train*, starting from the third coach all the way to the end, takes $\Delta t_2 = 20	ext{ s}$ to pass. If the train has a total of $N$ coaches, how many coaches just passed you? It's the $n^{	ext{th}}$ coach, the $(n+1)^{	ext{th}}$ coach, all the way to the $N^{	ext{th}}$ coach. The total number of coaches in this sequence is $(N - n + 1)$. Therefore, the total distance the train moved in this time is $(N - n + 1)L$. This gives us our second equation:
$$(N - n + 1)L = \frac{1}{2}a(\Delta t_2)^2$$

Now, we have two equations but three unknowns ($N$, $L$, and $a$). Here is where we use a classic physics trick: **division**. By dividing the second equation by the first, both the unknown length $L$ and the unknown acceleration $a$ cancel out completely!
$$\frac{(N - n + 1)L}{L} = \frac{\frac{1}{2}a(\Delta t_2)^2}{\frac{1}{2}a(\Delta t_1)^2}$$
$$N - n + 1 = \frac{\Delta t_2^2}{\Delta t_1^2}$$

Rearranging this to solve for the total number of coaches $N$, we get:
$$N = n + \left( \frac{\Delta t_2^2 - \Delta t_1^2}{\Delta t_1^2} ight)$$
Substituting the given values ($n=3$, $\Delta t_1=5$, $\Delta t_2=20$), we find that $N = 18$. The train has exactly 18 coaches.

### Part B: The Flash of the Last Coach

Now we need to find the time it took for *only* the very last coach to pass by. To do this, we can find the time $t'$ it took for all the coaches *except* the last one to pass.

The number of coaches before the last one is $(N - n)$. The distance they cover is $(N - n)L$. Using our master equation again:
$$(N - n)L = \frac{1}{2}a(t')^2$$

Once again, we divide this by our total distance equation to eliminate $L$ and $a$:
$$\frac{(t')^2}{\Delta t_2^2} = \frac{N - n}{N - n + 1}$$

We already know from Part A that $(N - n + 1) = \frac{\Delta t_2^2}{\Delta t_1^2}$, which means $(N - n) = \frac{\Delta t_2^2 - \Delta t_1^2}{\Delta t_1^2}$. Substituting these fractions into our ratio, the denominators cancel out, leaving us with:
$$(t')^2 = \Delta t_2^2 - \Delta t_1^2$$
$$t' = \sqrt{\Delta t_2^2 - \Delta t_1^2}$$

Finally, the time taken by the last coach is simply the total time $\Delta t_2$ minus the time $t'$ it took for the preceding coaches to pass:
$$\Delta t_{	ext{last}} = \Delta t_2 - \sqrt{\Delta t_2^2 - \Delta t_1^2}$$

Plugging in our numbers, we get $20 - \sqrt{20^2 - 5^2} = 20 - \sqrt{375} \approx 0.64	ext{ s}$.

Notice how the third coach took a leisurely 5 seconds to pass, but the last coach flashed by in just 0.64 seconds! This is the physical reality of constant acceleration—the longer the train moves, the faster it gets.

Visualized Solution

$u = 0, a = constant$

$S = \frac{1}{2} a t^{2}$

$L = \frac{1}{2} a (Δ t_{1})^{2}$

$(N - n + 1) L = \frac{1}{2} a (Δ t_{2})^{2}$

$N - n + 1 = \frac{( Δ t _{2} ) ^{2}}{( Δ t _{1} ) ^{2}}$

$N = n + (\frac{Δ t _{2}^{2} - Δ t _{1}^{2}}{Δ t _{1}^{2}})$

$(N - n) L = \frac{1}{2} a (t^{'})^{2}$

$t^{'} = Δ t_{2}^{2} - Δ t_{1}^{2}$

$Δ t_{last} = Δ t_{2} - t^{'}$

The Sigma Insight: Equations of Kinematics

The Setup

Part A

Part B

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u=0,a=constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S=21​at2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L=21​a(Δt1​)2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (N−n+1)L=21​a(Δt2​)2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N−n+1=(Δt1​)2(Δt2​)2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N=n+(Δt12​Δt22​−Δt12​​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (N−n)L=21​a(t′)2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t′=Δt22​−Δt12​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δtlast​=Δt2​−t′