Question

A train is travelling at a speed of 90 km h -1 . Brakes are applied so as to produce a uniform acceleration of – 0.5 m s -2 . Find how far the train will go before it is brought to rest.

Answer 1

Given that,

Acceleration a=−0.5m/s

2

Speed v=90km/h=25m/s

Using equation of motion,

v=u+at

Where,

v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the equation

Finally train will be rest so, final velocity,v=0

0=25−0.5t

25=0.5t

t=

0.5

25

​

t=50 sec

Again, using equation of motion,

S=ut+

2

1

​

at

2

Where, s = distance

v = final velocity

u = initial velocity

a = acceleration

t = time

Put the value into the equation

Where S is distance travelled before stop

s=25×50−

2

1

​

×0.5×(50)

2

s=625 m

So, the train will go before it is brought to rest is 625 m.

Answer 2

Given :-

A train is travelling at a speed of 90 km/hr . Brakes are applied so as to produce a uniform acceleration of - 0.5 m/s² .

Required to find :-

• Distance travelled before it is brought to rest ?

Equation used :-

1. v = u + at

2. s = ut + ½ at²

Solution :-

Given data :-

A train is travelling at a speed of 90 km/hr . Brakes are applied so as to produce a uniform acceleration of - 0.5 m/s² .

we need to find the distance travelled before it is brought to rest

So,

From the given data we can conclude that ;

• Initial velocity of the train ( u ) = 90 km/hr

• Acceleration ( a ) = - 0.5 m/s²

• Final velocity ( v ) = 0 km/hr

Since, the units of initial and velocity are in km/hr let's convert them to m/s because acceleration is given in m/s² .

We know that ;

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

So,

90 km/hr = ? m/s

=> 90 x 5/18

=> 5 x 5

=> 25 m/s

90 km/hr = 25 m/s

Similarly,

0 km/hr = 0 m/s

Hence,

• Initial velocity of the train ( u ) = 25 m/s

• Final velocity of the train ( v ) = 0 m/s

Using 1st equation of motion ;

✒ v = u + at

✒ 0 = 25 + ( - 0.5 ) ( t )

✒ 0 = 25 + ( - 0.5 t )

✒0 = 25 - 0.5t

✒ - 25 = - 0.5t

✒ - 0.5t = - 25

✒ 0.5t = 25

✒ t = 25/0.5

✒ t = 250/5

✒ t = 50 seconds

Hence,

Time taken ( t ) = 50 seconds

Using the 2nd equation of motion ;

s = ut + ½ at²

s = 25 x 50 + ½ x - 0.5 x 50 x 50

s = 1250 + ½ x - 0.5 x 2500

s = 1250 + ( - 0.5 x 1250 )

s = 1250 + ( - 625 )

s = 1250 - 625

s = 625 meters

Hence,

Displacement ( s ) = 625 meters

Remember, insome cases we can take displacement in terms of distance but we can't take distance in terms of speed .

Therefore ,

Distance travelled by the train before coming to rest is 625 meters