| Formula for converting from km/hr to m/s: $x$ km/hr =$ \left(x \times\dfrac{5}{18} \right) $m/s. |
| Therefore, Speed =$ \left(45 \times\dfrac{5}{18} \right) $m/sec=$ \dfrac{25}{2} $m/sec. |
Total distance to be covered = (360 + 140) m = 500 m.
| $\therefore$ Required time =$ \left(\dfrac{500 \times 2}{25} \right) $sec= 40 sec. |
Let us name the trains as A and B. Then,
A's speed : B's speed = $ \sqrt{b} $ : $ \sqrt{a} $ = $ \sqrt{16} $ : $ \sqrt{9} $ = 4 : 3.
| 2 kmph =$ \left(2 \times\dfrac{5}{18} \right) $m/sec =$ \dfrac{5}{9} $m/sec. |
| 4 kmph =$ \left(4 \times\dfrac{5}{18} \right) $m/sec =$ \dfrac{10}{9} $m/sec. |
Let the length of the train be $ x $ metres and its speed by $ y $ m/sec.
| Then, | $ x $ | = 9 and | $ x $ | = 10. |
$\therefore 9 y $ - 5 = $ x $ and 10$\left(9 y - 10\right)$ = 9$ x $
$\Rightarrow 9 y $ - $ x $ = 5 and 90$ y $ - 9$ x $ = 100.
On solving, we get: $ x $ = 50.
$\therefore$ Length of the train is 50 m.
| Relative speed = | = (45 + 30) km/hr |
| =$ \left(75 \times\dfrac{5}{18} \right) $m/sec | |
| =$ \left(\dfrac{125}{6} \right) $m/sec. |
We have to find the time taken by the slower train to pass the DRIVER of the faster train and not the complete train.
So, distance covered = Length of the slower train.
$\therefore$ Distance covered = 500 m
| $\therefore$ Required time = $\left(500 \times\dfrac{6}{125} \right) $= 24 sec. |
Let the speed of the second train be $ x $ km/hr.
| Relative speed | = $\left( x + 50\right)$ km/hr |
| =$\left[\left( x + 50\right) \times \dfrac{5}{18}\right] $sec | |
| =$\left[\dfrac{250 + 5x}{18}\right]$sec. |
Distance covered = (108 + 112) = 220 m.
| $\therefore \frac{220}{\left(\dfrac{250 + 5x}{18} \right) } $= 6 |
$\Rightarrow 250 + 5 x $ = 660
$\Rightarrow x$ = 82 km/hr.
| Speed =$ \left(54 \times\dfrac{5}{18} \right) $m/sec = 15 m/sec. |
Length of the train = (15 \times 20)m = 300 m.
Let the length of the platform be $ x $ metres.
| Then,$ \dfrac{x + 300}{36} $= 15 |
$\Rightarrow x $ + 300 = 540
$\Rightarrow x $ = 240 m.
| Speed =$ \left(72 \times\dfrac{5}{18} \right) $m/sec= 20 m/sec. |
Time = 26 sec.
Let the length of the train be $ x $ metres.
| Then,$ \dfrac{x + 250}{26} $= 20 |
$\Rightarrow x $ + 250 = 520
$\Rightarrow x $ = 270.
Let the speed of the slower train be $ x $ m/sec.
Then, speed of the faster train = 2$ x $ m/sec.
Relative speed = $\left( x + 2 x\right)$ m/sec = 3$ x $ m/sec.
| $\therefore \dfrac{(100 + 100)}{8} $= 3$ x $ |
$\Rightarrow$ 24$ x $ = 200
| $\Rightarrow x $ =$ \dfrac{25}{3} $. |
| So, speed of the faster train =$ \dfrac{50}{3} $m/sec |
| =$ \left(\dfrac{50}{3} \times\dfrac{18}{5} \right) $km/hr |
= 60 km/hr.
| 4.5 km/hr =$ \left(4.5 \times\dfrac{5}{18} \right) $m/sec =$ \dfrac{5}{4} $m/sec = 1.25 m/sec, and |
| 5.4 km/hr =$ \left(5.4 \times\dfrac{5}{18} \right) $m/sec =$ \dfrac{3}{2} $m/sec = 1.5 m/sec. |
Let the speed of the train be $ x $ m/sec.
Then, $\left(x - 1.25\right)$ x 8.4 = $\left( x - 1.5\right)$ x 8.5
$\Rightarrow 8.4 x $ - 10.5 = 8.5$ x $ - 12.75
$\Rightarrow 0.1 x $ = 2.25
$\Rightarrow x $ = 22.5
| $\therefore$ Speed of the train =$ \left(22.5 \times\dfrac{18}{5} \right) $km/hr = 81 km/hr. |
Let the length of the first train be $ x $ metres.
| Then, the length of the second train is$ \left(\dfrac{x}{2} \right) $metres. |
| Relative speed = (48 + 42) kmph =$ \left(90 \times\dfrac{5}{18} \right) $m/sec = 25 m/sec. |
| $\therefore \dfrac{[x + (x/2)]}{25} $= 12 or$ \dfrac{3x}{2} $= 300 or $ x $ = 200. |
$\therefore$ Length of first train = 200 m.
Let the length of platform be $ y $ metres.
| Speed of the first train =$ \left(48 \times\dfrac{5}{18} \right) $m/sec =$ \dfrac{40}{3} $m/sec. |
| $\therefore (200 + y ) \times \dfrac{3}{40} $= 45 |
$\Rightarrow 600 + 3 y $ = 1800
$\Rightarrow y $ = 400 m.
| Speed =$ \left(78 \times\dfrac{5}{18} \right) $m/sec=$ \left(\dfrac{65}{3} \right) $m/sec. |
Time = 1 minute = 60 seconds.
Let the length of the tunnel be $ x $ metres.
| Then,$ \left(\dfrac{800 + x}{60} \right) $=$ \dfrac{65}{3} $ |
$\Rightarrow 3\left(800 + x \right)$ = 3900
$\Rightarrow x $ = 500.
| Speed=$ \left(60 \times\dfrac{5}{18} \right) $m/sec=$ \left(\dfrac{50}{3} \right) $m/sec. |
| Length of the train = $(Speed \times Time) = \left(\dfrac{50}{3} \times 9\right) $m = 150 m. |
Suppose they meet $ x $ hours after 7 a.m.
Distance covered by A in $ x $ hours = 20$ x $ km.
Distance covered by B in $\left( x - 1\right)$ hours = 25$\left( x - 1\right)$ km.
$\therefore 20 x $ + 25$\left( x - 1\right)$ = 110
$\Rightarrow 45 x $ = 135
$\Rightarrow x $ = 3.
So, they meet at 10 a.m.
| Speed =$ \left(45 \times\dfrac{5}{18} \right) $m/sec=$ \left(\dfrac{25}{2} \right) $m/sec. |
Time = 30 sec.
Let the length of bridge be $ x $ metres.
| Then,$ \dfrac{130 + x}{30} $=$ \dfrac{25}{2} $ |
$\Rightarrow$ 2$\left(130 + x\right)$ = 750
$\Rightarrow x $ = 245 m.
Let the speed of each train be $ x $ m/sec.
Then, relative speed of the two trains = 2$ x $ m/sec.
| So, 2$ x $ =$ \dfrac{(120 + 120)}{12} $ |
$\Rightarrow 2 x $ = 20
$\Rightarrow x $ = 10.
| $\therefore$ Speed of each train = 10 m/sec =$ \left(10 \times\dfrac{18}{5} \right) $km/hr = 36 km/hr. |