Question

a train travelling at 72km/h is to be bought to rest at 200m then its acceleration is to be

Answer 1

Explanation:

given,

u=72 kmph= 20m/s

v=0m/s

distance traveled is 200 m

from third equation of motion

v²-u²=2as

a= 0²-20²/2(200)

a=-400/400

a= -1 m/sec²

-ve sign of acceleration show that it it is regarding

Answer 2

Answer:

Retardation of the train is -1 m/s

Explanation:

Given :-

Initial velocity (u) = 72 Km/h

Final velocity (v) = 0 Km/h

Distance (s) = 200 m

To find :

Retardation of the train (a)

Solution :-

u = 72 Km/h

• 72 × (5/18) m/s
• 20 m/s

v = 0 m/s

s = 200 m

Using the third equation of motion :

⇒ v² - u² = 2as

• (0)² - (20)² = 2 × a × 200
• 0 - 400 = 2a × 200
• -400 = 400a
• a = -400/400
• a = -1

∴ Retardation of the train is -1 m/s and retardation is negative acceleration, if a body moves it is called as acceleration and if the body slows down it is called retardation.