Question

26. 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 Find how far the train will go before it is brought to rest.​

Answer 1

Answer :-

625

Explanation :-

Given :

• Initial velocity of speed,u => 90km/hr = 25m/s   [90km/hr = (90*1000)/(3600)]

• Final velocpity of the speed,v => 0

• Acceleration,a => 0.5m/s^2

To Find :

• Distance travelled,s => ?

Solution :

According to the second equation of motion,

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

$\sf{}:\implies 0=25^2+2\times 0.5\times s$

$\sf{}:\implies 0=625+2\times 0.5\times s$

$\sf{}:\implies -625=1.0\times s$

$\sf{}\therefore s=625m$

Therefore,distance travelled is equal to 625m.

☆Know more☆

Equations of Motion:

Relation among velocity, distance, time and acceleration is called equation of motion. There are three equations of motion:

• 1st Equation of motion ( or Velocity-time relation): v = u + at

• 2nd Equation of motion ( or Position-time relation): s = ut + 1/2 at^2

• 3rd Equation of motion ( or Position- velocity relation): v^2 - u^2 = 2as