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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.

User Lashonda
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✴ 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.

User Madan Ram
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