Let the speeds of the two trains be $ x $ m/sec and y m/sec respectively.
Then, length of the first train = 27$ x $ metres,
and length of the second train = 17$ y $ metres.
| $\therefore \dfrac{27x + 17y}{x+ y} $= 23 |
$\Rightarrow$ 27$ x $ + 17$ y $ = 23$ x $ + 23$ y $
$\Rightarrow$ 4$ x $ = 6$ y $
| $\Rightarrow \dfrac{x}{y} $=$ \dfrac{3}{2} $. |
Let x is the length of the train and v is the speed
Time taken to move the post = 8 s
=> x/v = 8
=> x = 8v --- (1)
Time taken to cross the platform 264 m long = 20 s
(x+264)/v = 20
=> x + 264 = 20v ---(2)
Substituting equation 1 in equation 2, we get
8v +264 = 20v
=> v = 264/12 = 22 m/s
= 22×36/10 km/hr = 79.2 km/hr
| Speed of the train relative to man =$ \left(\dfrac{125}{10} \right) $m/sec |
| =$ \left(\dfrac{25}{2} \right) $m/sec. |
| =$ \left(\dfrac{25}{2} \times\dfrac{18}{5} \right) $km/hr |
= 45 km/hr.
Let the speed of the train be $ x $ km/hr. Then, relative speed = $\left( x - 5\right)$ km/hr.
$\therefore x $ - 5 = 45 $\Rightarrow$ $ x $ = 50 km/hr.
Relative speed = (60+ 90) km/hr
| =$ \left(150 \times\dfrac{5}{18} \right) $m/sec |
| =$ \left(\dfrac{125}{3} \right) $m/sec. |
Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.
| Required time =$ \left(2000 \times\dfrac{3}{125} \right) $sec = 48 sec. |
| Speed =$ \left(\dfrac{300}{18} \right) $m/sec =$ \dfrac{50}{3} $m/sec. |
Let the length of the platform be $ x $ metres.
| Then,$ \left(\dfrac{x + 300}{39} \right) $=$ \dfrac{50}{3} $ |
$\Rightarrow 3\left(x + 300\right)$ = 1950
$\Rightarrow x $ = 350 m.
Let the length of the train be $ x $ metres and its speed by $ y $ m/sec.
| Then,$ \dfrac{x}{y} $= 8 $\Rightarrow x = 8 y $ |
| Now,$ \dfrac{x + 264}{20} $= $ y $ |
$\Rightarrow 8 y $ + 264 = 20$ y $
$\Rightarrow y $ = 22.
| $\therefore Speed = 22 m/sec$ =$ \left(22 \times\dfrac{18}{5} \right) $km/hr = 79.2 km/hr. |
Relative speed = (120 + 80) km/hr
| =$ \left(200 \times\dfrac{5}{18} \right) $m/sec |
| =$ \left(\dfrac{500}{9} \right) $m/sec. |
Let the length of the other train be $ x $ metres.
| Then,$ \dfrac{x + 270}{9} $=$ \dfrac{500}{9} $ |
$\Rightarrow x $ + 270 = 500
$\Rightarrow x $ = 230.
Let the speed of the trains be $x$ and $y$ respectively
length of train1 = 27$x$
length of train2 = 17$y$
Relative speed= $x+ y$
Time taken to cross each other = 23 s
=> $\left (27x + 17 y\right)$/$\left(x+y\right)$ = 23
=> $\left (27x + 17 y\right)$/ = 23$\left(x+y\right)$
=> $4x = 6y$
=> $x/y = 6/4 = 3/2$
Let the length of the train be $ x $ metres and its speed by $ y $ m/sec.
| Then,$ \dfrac{x}{y} $= 15 $\Rightarrow$ $ y $ =$ \dfrac{x}{15} $. |
| $\Rightarrow \dfrac{x + 100}{25} $=$ \dfrac{x}{15} $ |
$\Rightarrow 15( x $ + 100) = 25$ x $
$\Rightarrow 15 x $ + 1500 = 25$ x $
$\Rightarrow 1500 = 10 x $
$\Rightarrow x $ = 150 m.
Length of the train = distance covered in crossing the post = speed × time = speed × 18
Speed of the train = 300/18 m/s = 50/3 m/s
Time taken to cross the platform = 39 s
$\left(300+x\right)$/(50/3) = 39 s where x is the length of the platform
300+x = (39 × 50) / 3 = 650 meter
x = 650-300 = 350 meter