Guys pls help with this question
a train is travelling at a speed of 54 kmph brakes are applied so as to produce a uniform acceleration of -0.25 ms. Calculate the distance travelled by the train before it comes to rest.
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Answers
Answer:
Answer:−
✰ distance travelled by the train = 450m.
\large{\underline{\bf{\blue{Explanation:-}}}}
Explanation:−
✰ we used the equation v² - u² = 2as
where,
➜⠀⠀⠀v = final velocity
➜⠀⠀⠀u = initial velocity
➜⠀⠀⠀a = acceleration
➜⠀⠀⠀s = distance
\large{\underline{\bf{\green{Given:-}}}}
Given:−
✰ ⠀⠀⠀Inital velocity (u) = 54km/h
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 54×(5/18)
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 15m/s
✰⠀⠀⠀ Final velocity (v) = 0
✰ ⠀⠀⠀Acceleration (a) = -(0.25)
\large{\underline{\bf{\green{To\:Find:-}}}}
ToFind:−
➜⠀⠀⠀we need to find the distance ⠀⠀⠀⠀⠀⠀travelled by the train.
\huge{\underline{\bf{\red{Solution:-}}}}
Solution:−
➩⠀⠀⠀v² - u² = 2as
putting all values in the equation.
we get ,
➩⠀⠀⠀0² - (15)² = 2 × (-0.25)×s
➩⠀⠀⠀-225 = - 0.5s
➩⠀⠀⠀s = 225/0.5
➩⠀⠀⠀s = 450m
So the distance travelled by the train is 450 m.
• Initial velocity, u = 54km/h⠀⠀⠀
Converting 54 km/h to m/s :-⠀⠀⠀⠀
( Divide initial velocity by 3.6)
Hence, initial velocity = 15 m/s
• Final velocity, v = 0 m/s
• Acceleration , a = -0.25m/s²
• Distance travelled by the train
To find distance, we need to use third equation of motion.
v² - u² = 2as
Where,
v = Final velocity
u = Initial velocity
a = Acceleration
s = Distance traveled
Now, put the given values in the equation.
⟹ v² - u² = 2as
⟹ 0² - (15)² = 2 × (-0.25)×s
⟹ -225 = - 0.5s
⟹ s = 450m
Hence,
the distance travelled by the train is 450 m
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Know more :-
• v = u + at (First equation of motion)
• s = ut + ½ at² (Second equation of motion)
• v² = u² + 2as (third equation of motion)