Question

A train is travelling at a speed of 80 kmh-1

. Brakes are applied so as to produce a

uniform retardation of 0.5 ms
-2
. Find how far the train goes before it stops.

correct answer means brainliest

incorrect answer means delete​

Answer 1

Answer:

Given that,

Acceleration a=0.5m/s

Speed v=80km/h=22.5m/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=22.5+0.5t

22.5=0.5t

t= 22.5/0.5

t=45 sec

Again, using equation of motion,

S=ut+ 1/2atsquare

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=22.5×45-1/2x0.5x45 square

s=1012.5-506.25

s=506.25 m

So, the train will go before it is brought to rest is 506.25m.

Answer 2

Solution :

As per the given data ,

• Initial velocity = 80 km / hr = 80 x 5 / 18 = 22.2 m /s (approx. )

[ in order to convert it into m /s multiply by 5 / 18 ]

• Retardation (a) = 0.5 m/s ²
• Final velocity (v) = 0 m/s [ stops ]

As the train is moving with uniform acceleration throughout it's motion we can use the third equation in order to find the distance traveled by the train .

By using third equation of motion ,

$\dag \boxed{\mathtt{v^2 = u^2 + 2as } }$

On rearranging ,

⇒ s = v² - u² / 2a

Now let's substitute the given values in the above equation ,

⇒ s = - 80 x 80 /2 x -0.5

⇒ s = 0 - 6400 / - 1

⇒ s = 6400 m

Hence ,

The distance traveled by the train is 6400 m.