Question

A train is travelling at speed 90 km per hour . Breaks are applied so as to produce uniform acceleration of -0.6 metre per second ^2. Find how far train will go before it is brought to rest​

Answer 1

Given :

➳ Initial speed of train = 90kmph

➳ Acceleration of train = -0.6m/s²

(-ve sign shows retardation.)

To Find :

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

Concept :

➛ Applied brakes exert retarding force on the train and it results into uniform retardation in motion.

➛ Distance covered by moving object before coming to rest is called as stopping distance.

✭ Third equation of kinematics :

• v² - u² = 2as

where,

◕ v denotes final velocity

◕ u denotes initial velocity

◕ a denotes acceleration

◕ s denotes distance

Conversion :

➾ 1kmph = 5/18mps

➾ 90kmph = 90 × 5/18 = 25mps

Calculation :

$\longrightarrow\tt\:v^2-u^2=2as\\ \\ \longrightarrow\tt\:(0)^2-(25)^2=2(-0.6)s\\ \\ \longrightarrow\tt\:s=\dfrac{-625}{-1.2}\\ \\ \longrightarrow\underline{\boxed{\tt{s=520.8m}}}$

Answer 2

Answer:

520.834 m

Explanation:

u = 90kmph = 90×(5÷18) = 25m/s

acceleration = -0.6m/s²

v = 0 (since it goes to rest)

using the formula v²-u²=2as,

substitute the above values,

(0)²-(25)² = 2(-0.6)s

-625 = -1.2s

s = (-625)÷(-1.2)

s = 520.834

hope this helps !