Question

A train is travelling at a speed of 90 km per hour brakes are applied so as to produce a uniform acceleration of - 0.5 metre per second square find how far the train will go before it is brought to rest rest

Answer 1

Answer:

u=90kmph

v=0

a= -0.5mps

from 1 law of motion

v=u+at

0=90 *-0.5*t

0.5t=90

t=180s

from 2 law of equation

s=ut+1/2at2

s=90*180+1/2*-0.5*180*180

after division you will get the answer

Answer 2

The distance covered by train before it is brought to rest is 625 meters.

Explanation:

Initial speed of train is 90 km/h or 25 m/s

Finally it comes to rest, v = 0

Acceleration, $a=-0.5\ m/s^2$

Let d is the distance covered by train before it is brought to rest. Now using third equation of motion to find it as :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0-(25)^2}{2\times (-0.5)}\\\\d=625\ m$

Learn more,

Kinematics

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