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)

Question

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)

2 Answers

Impact of this question

2070 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-09T00:43:13

$0.65 s$ $0.65" s"$ , rounded to one decimal place.

Explanation:

Let the length of each carriage be $L m$ $L" m"$
Suppose that train accelerates at a constant rate of acceleration $a$ $a$

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

Inserting various values for the first carriage we get
$L = 0 \times 4 + \frac{1}{2} a \times 4^{2}$ $L=0xx4+1/2axx4^2$
$\Rightarrow \frac{1}{2} a \times 4^{2} = L$ $=>1/2axx4^2=L$
$\Rightarrow a = L \times \frac{2}{16}$ $=>a=Lxx2/16$
$\Rightarrow a = \frac{L}{8} {ms}^{- 2}$ $=>a=L/8" ms"^-2$

From (1)
Time taken to pass $9$ $9$ carriages through reference point
$9 L = 0 \times 4 + \frac{1}{2} \frac{L}{8} t_{9}^{2}$ $9L=0xx4+1/2L/8t_9^2$
$\Rightarrow \frac{1}{16} t_{9}^{2} = 9$ $=>1/16t_9^2=9$
$\Rightarrow t_{9} = \sqrt{144} = 12 s$ $=>t_9=sqrt144=12s$
Similarly time taken to pass $10$ $10$ carriages through reference point
$10 L = 0 \times 4 + \frac{1}{2} \frac{L}{8} t_{10}^{2}$ $10L=0xx4+1/2L/8t_10^2$
$\Rightarrow \frac{1}{16} t_{10}^{2} = 10$ $=>1/16t_10^2=10$
$\Rightarrow t_{10} = \sqrt{160} = 12.65 s$ $=>t_10=sqrt160=12.65s$ , rounded to two decimal places

Therefore time taken by tenth carriage to pass through the reference point $= t_{10} - t_{9} = 0.65 s$ $=t_10-t_9=0.65s$ , rounded to two decimal places

Answer 2 · 2017-02-10T20:34:45

$≅$ $~=$ 0.65

Explanation:

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

t= $\sqrt{\frac{2 \cdot L}{a}}$ $sqrt((2*L)/a)$
$t_{9}$ $t_9$ = $sqrt((2*9*2cancelu*4)/cancelu)$ =12sec
$t_10$ = $sqrt((2*10*2cancelu*4)/cancelu)$ = $sqrt160$ $~=$ 12.65

Δt $~=$ 0.65

Science

Math

Humanities

... and beyond

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)

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question