Question

A train is travelling at a speed of 70km/h..Brakes are applied so as to produce a uniform accelerationof -0.5m/s²... find how far the train will go before it is brought to rest.​

Answer 1

Given-

☞Initial speed of train of the train= 70km/h

☞Final speed = 0 (Brakes applied)

☞Deceleration = 0.5m/s²

 

To Find-

Distance Covered by the train-?

 

Formula to be used-

☞ $\boxed{\sf{ v^2-u^2 = 2as}}$

where,

v= final speed

u= initial speed

a= acceleration

s= distance covered

 

Solution-

☞At first change km/h to m/s.

we know

1 km = 1000 m

1 hr= 3600 s

so,

$\sf{1km/h = \dfrac{1000m}{3600s} = \dfrac{5}{18} m/s }$

from this

$\sf{\implies 70km/h = (\dfrac{5}{18} \times 70)m/s }$

$\bf{\implies u = 19.44 m/s}$

☞Now By using 3rd equation of motion, i.e

$\bf{ v^2-u^2 = 2as}$

Put given values

$\sf{\implies 0^2 - (19.44)^2 = 2 \times (-0.5) \times s}$

$\sf{\implies -377.91 = -1 \times s}$

$\bf{ s= 377.91 m}$

 

Therefore, The distance covered by the train is 377.91m.

 

Answer 2

hence the distance covered is 200m