Question

Brakes are applied to train travelling at 72 km/h after passing over 200m . its velocity is reduced to 36 km/h find the retardation

Answer 1

Given :-

Brakes are applied to train travelling 72 km/hr after passing over 200 meters it's velocity got reduced to 36 km/hr .

Required to find :-

• Retardation of the train

Equation used :-

• v² - u² = 2as

Solution :-

Given information :-

Brakes are applied to train travelling 72 km/hr after passing over 200 meters it's velocity got reduced to 36 km/hr .

we need to find the retardation of the train .

From the given information wr can conclude that ;

• Initial velocity of the train = 72 km/hr

• Final velocity of the train = 36 km/hr

• Distance travelled = 200 meters

Here,

We need to convert the speed from km/hr into m/s

So, Using the formula

$\boxed {\tt \red{1 \: kmh {r}^{ - 1} = \frac{5}{18} \: m {s}^{ - 1} }}$

Hence,

72 km/hr = ? m/s

=> 5/18 x 72

=> 5 x 4

=> 20 m/s

Similarly,

36 km/hr = ? m/s

=> 5/18 x 36

=> 5 x 2

=> 10 m/s

This implies ;

• Initial velocity of the train = 20 m/s

• Final velocity of the train = 10 m/s

Now,

Using the 3rd equation of motion ;

v² - u² = 2as

( 10 )² - ( 20 )² = 2 x a x 200

100 - 400 = 2a x 200

- 300 = 2a x 200

- 300/2 = a x 200

- 150 = 200a

=> 200a = - 150

=> a = - 150/200

=> a = - 3/4 m/s²

=> a = - 0.75 m/s²

Therefore ;

Retardation of the train = - 0.75 m/s²

Point to remember :-

• Negative acceleration is retardation

Answer 2

Answer:

Given :-

Brakes are applied to train travelling 72 km/hr after passing over 200 meters it's velocity got reduced to 36 km/hr .

Required to find :-

Retardation of the train

Equation used :-

v² - u² = 2as

Solution :-

Given information :-

Brakes are applied to train travelling 72 km/hr after passing over 200 meters it's velocity got reduced to 36 km/hr .

we need to find the retardation of the train .

From the given information wr can conclude that ;

Initial velocity of the train = 72 km/hr

Final velocity of the train = 36 km/hr

Distance travelled = 200 meters

Here,

We need to convert the speed from km/hr into m/s

So, Using the formula

\boxed {\tt \red{1 \: kmh {r}^{ - 1} = \frac{5}{18} \: m {s}^{ - 1} }}

1kmhr

−1

=

18

5

ms

−1

Hence,

72 km/hr = ? m/s

=> 5/18 x 72

=> 5 x 4

=> 20 m/s

Similarly,

36 km/hr = ? m/s

=> 5/18 x 36

=> 5 x 2

=> 10 m/s

This implies ;

Initial velocity of the train = 20 m/s

Final velocity of the train = 10 m/s

Now,

Using the 3rd equation of motion ;

v² - u² = 2as

( 10 )² - ( 20 )² = 2 x a x 200

100 - 400 = 2a x 200

- 300 = 2a x 200

- 300/2 = a x 200

- 150 = 200a

=> 200a = - 150

=> a = - 150/200

=> a = - 3/4 m/s²

=> a = - 0.75 m/s²

Therefore ;

Retardation of the train = - 0.75 m/s²

Point to remember :-

Negative acceleration is retardation