Question

A train is moving at a speed of 90km/h. After applying brakes
a o.5 m/s² acceleration is produced. Find how far distance will the train go before it come to rest.​

Answer 1

Answer:

A train is moving at a speed of 90km/h. After applying brakes

a o.5 m/s² acceleration is produced. Find how far distance will the train go before it come to rest.

Explaination

Given  u = 90 kmph 

               = 90 * 5/18 = 25 m/s

            v = 0.

           a = - 2.5 m/s²

We have the equation of motion:

    v² = u² + 2 a s 

    0 = 25² - 2 * 2.5 * s

    s = 125 meters.

So the train travels 125 meters before it comes to a halt.

Answer 2

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.