Question

A train is travelling at a speed of 60 km/h.brakes are applyed as to produce a uniform acceleration of -0.5 m/s² . find how far will the train go before it is brought to rest .​

Answer 1

Given:-

→Initial velocity of the train = 60km/h

→Acceleration of the train = -0.5m/s²

To find:-

→Distance the train will cover before coming to rest

Solution:-

•Final velocity of the train will be zero as it finally comes to rest after application of brakes.

Firstly,let's convert the initial velocity of the train from km/h to m/s.

=>1km/h = 5/18m/s

=>60km/h = 5/18×60

=>50/3 m/s

By using the 3rd equation of motion,we get:-

=>v²-u² = 2as

=>0-(50/3)² = 2(-0.5)s

=> -2500/9 = -s

=>s = 2500/9

=>s = 277.8m. (approximately)

Thus,the train will cover a distance of 277.8m before coming to rest.

Some Extra Information:-

The 3 equations of motion for a body moving with uniform acceleration are:-

•v = u+at

•s = ut+1/2at²

•v²-u² = 2as

Answer 2

Answer :-

278.89m

Explanation :-

Given :

Speed,v = 60km/hr

Convert it into m/s

$\sf{}\Rightarrow \dfrac{(60\times 1000)m}{(1\times 60\times 60)s}$

$\sf{}\Rightarrow \dfrac{60000m}{3600s}$

$\sf{}\Rightarrow \dfrac{50}{3}m/s$

$\sf{}\Rightarrow 16.7m/s$

Acceleration,a = -0.5m/s^2

Initial velocity,u = 0 m/s [As the train is going to stop]

To Find :

Distance,s = ?

Solution :

According to the third equation of motion,

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

Put their values and find “s”

$\implies\sf{}(0)^2=(16.7)^2+2\times -0.5\times s$

$\implies \sf{}0=278.89-1.0\times s$

$\implies \sf{}-278.89=-1 \times s$

$\implies \sf{}\dfrac{-278.89}{-1}=s$

$\sf{}\therefore s= 278.89m$

Therefore,distance travelled is equal to 278.89m. (approx)

☆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 + 1/2at^2

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