Question

A train is traveling at a speed of 90 km h-¹.braked are applied so as to produce a uniform acceleration of-0.5m s-².find how far te train will go before it is brought to rest

Answer 1

Given :

▪ Initial speed of train = 90kmph

▪ Retardation = -0.5m/s²

To Find :

▪ Distance covered by train before it is brought to rest.

Concept :

↗ Since, acceleration of train has said to be constant throughout the motion, we can apply equation of kinematics to solve this question.

↗ This question is completely based on the concept of stopping distance.

✴ Third equation of kinematics :

➡ v² - u² = 2as

Conversion :

❇ 1kmph = 5/18mps

❇ 90kmph = 90×5/18 = 25mps

CalculaTion :

→ v² - u² = 2as

→ (0)² - (25)² = 2(-0.5)s

→ -(625) = -(1)s

→ s = 625m

Answer 2

Given ,

Initial speed of train = 90 km/h = 90*(5/18) = 25 m/s

Retardation , a = - 0.5 m/s²

Distance , s = ? m

Final speed , v = 0 m/s

Use 3rd equation of motion to find the distance covered by train before going to rest.

$\bold{\bf{\red{v^2-u^2=2as}}}\\\\\implies \bold{o^2-25^2=2(-0.5)s}\\\\\implies \bold{-625=-s}\\\\\implies \bold{\bf{\red{s=625\;meters}}}$

So train will move 625 meters before going to rest .