# A train previously at rest starts accelerating. It takes 4 seconds for the first carriage to cross the reference point K. How long will it take for the 10th carriage to cross the same point? (All carriages are the same width)

Feb 9, 2017

$0.65 \text{ s}$, rounded to one decimal place.

#### Explanation:

Let the length of each carriage be $L \text{ m}$
Suppose that train accelerates at a constant rate of acceleration $a$

We use the kinematic equation
$s = u t + \frac{1}{2} a {t}^{2}$ .......(1)
where $s$ is displacement, $u$ initial velocity and $t$ is time of travel

Inserting various values for the first carriage we get
$L = 0 \times 4 + \frac{1}{2} a \times {4}^{2}$
$\implies \frac{1}{2} a \times {4}^{2} = L$
$\implies a = L \times \frac{2}{16}$
$\implies a = \frac{L}{8} {\text{ ms}}^{-} 2$

From (1)
Time taken to pass $9$ carriages through reference point
$9 L = 0 \times 4 + \frac{1}{2} \frac{L}{8} {t}_{9}^{2}$
$\implies \frac{1}{16} {t}_{9}^{2} = 9$
$\implies {t}_{9} = \sqrt{144} = 12 s$
Similarly time taken to pass $10$ carriages through reference point
$10 L = 0 \times 4 + \frac{1}{2} \frac{L}{8} {t}_{10}^{2}$
$\implies \frac{1}{16} {t}_{10}^{2} = 10$
$\implies {t}_{10} = \sqrt{160} = 12.65 s$, rounded to two decimal places

Therefore time taken by tenth carriage to pass through the reference point$= {t}_{10} - {t}_{9} = 0.65 s$, rounded to two decimal places

Feb 10, 2017

$\cong$0.65

#### Explanation:

α= $\frac{u}{4}$
u is the speed gained in 4 secs
L=${u}^{2} / \left(2 a\right)$ $\Leftrightarrow$ L=2u
L is the length of a carriage

t=$\sqrt{\frac{2 \cdot L}{a}}$
${t}_{9}$= $\sqrt{\frac{2 \cdot 9 \cdot 2 \cancel{u} \cdot 4}{\cancel{u}}}$=12sec
${t}_{10}$=$\sqrt{\frac{2 \cdot 10 \cdot 2 \cancel{u} \cdot 4}{\cancel{u}}}$ =$\sqrt{160}$$\cong$12.65

Δt$\cong$0.65