Question

A train is moving with a speed of 90 km/h . On applying brakes a retardation of 1.25
m/s² in caused . At what distance the driver should applying brakes to stop the train at the station. ​

Answer 1

Given :-

• Initial velocity (u) = 90 km/h = 25 m/s
• Final velocity (v) = 0 m/s (as it applied brakes)
• Retardation = -1.25 m/s²

To find :-

• Distance (s)

According to the question,

By using Newtons third equation of motion,

⇒ v² = u² + 2as

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• s = Distance

⇒ Substituting the values,

⇒ (0)² = (25)² + 2 × (-1.25) × s

⇒ 0 = 625 + (-2.5)s

⇒ 0 - 625 = -2.5s

⇒ -625 = -2.5s

⇒ 625 ÷ 2.5 = s

⇒ 250 = s

.°. The distance is 250 m..

_________________________________

Answer 2

$\begin{lgathered}\begin{lgathered}\begin{lgathered}\tt {\pink{Given}}\begin{cases} \sf{\green{Initial velocity = 25 m/s}}\\ \sf{\blue{Final velocity = 0 m/s}}\\ \sf{\orange{Retardation = 1.25 </p><p>m/s^2}}\end{cases}\end{lgathered} \:\end{lgathered}\end{lgathered}$

To find:-

• Distance

Solution:-

⟹ (0)² = (25)² + 2 × (-1.25) × s

⟹ 0 = 625 + (-2.5)s

⟹ 0 - 625 = -2.5s

⟹ -625 = -2.5s

⟹ 625 ÷ 2.5 = s

⟹ $\boxed{s= 250}$