✴ A train is uniformly slowed down from 35m/s to 21m/s over a distance of 350m. Calculate the acceleration and the time taken to come to rest from the 35m/s.
✒ The acceleration is - 1.12 ms-² and the time taken to come to rest from the 35m/s is 31.3 seconds .
Given :-
A train is uniformly slowed down from 35m/s to 21m/s over a distance of 350m.
To Find :-
Calculate the acceleration and the time taken to come to rest from the 35m/s.
Calculation :-
According to the question,
Initial velocity ( u ) = 35 m/s
Final velocity ( v ) = 21 m/s
Displacement ( s ) = 350 meters
♠ A train is uniformly slowed down from 35m/s to 21m/s over a distance of 350m.
From the equation of motion ,
↗ v² - u² = 2as
➡ ( 21 )² - ( 35 )² = 2 x a x 350
➡ 441 - 1225 = 2 x a x 350
➡ - 784 = 700 x a
➡ - 784 = 700a
➡ 700a = - 784
➡ a = -784/700
➡ a = - 1.12 ms-²
So, The Acceleration of the train is - 1.12 ms-².
Initial velocity ( u ) = 35 m/s
Acceleration ( a ) = - 1.12 m/s²
Final velocity ( v ) = 0 m/s
From, the equation of motion ;
↗ v = u + at
➡ 0 = 35 + ( - 1.12 x t )
➡ 0 = 35 + ( - 1.12t )
➡ 0 - 35 = - 1.12t
➡ - 35 = - 1.12t
➡ t = 35/1.12
➡ t = 3500/112
➡ t = 31.3 seconds
The time taken to come to rest from the 35m/sis 31.3seconds.