Question

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.

Answer 1

ANSWER

Given : -

A train is uniformly slowed down from 35m/s to 21m/s over a distance of 350m.

Required to find : -

• Acceleration ?

• Time taken to come to rest from 35 m/s ?

Equations used : -

❶ v² - u² = 2as

❷ v = u + at

Here,

• v = Final velocity

• u = Initial velocity

• a = acceleration

• t = time taken

• s = displacement

sOluTioN : -

A train is uniformly slowed down from 35m/s to 21m/s over a distance of 350m.

We need to find the ;

• Acceleration ?

• Time taken to come to rest from 35 m/s ?

So,

From the given information we can conclude that ;

Initial velocity of the train ( u ) = 35 m/s

Final velocity of the train ( v ) = 21 m/s

Displacement ( s ) = 350 meters

Now,

Let's find the acceleration of the train !

Using the equation of motion ;

i.e. 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 m/s²

Hence,

➔ Acceleration of the train ( a ) = - 1.12 m/s²

Now,

Let's solve the 2nd part of the Question .

We were asked to find the time taken by the train to come to rest from 35 m/s .

So,

Since it is mentioned that ;

The acceleration throughout the journey is constant .

This Implies ;

Initial velocity of the train ( u ) = 35 m/s

Acceleration of the train ( a ) = - 1.12 m/s²

Final velocity of the train ( v ) = 0 m/s

Now,

Let's find the time taken by the train .

So,

Using the equation of motion ;

i.e. v = u + at

➔ 0 = 35 + ( - 1.12 x t )

➔ 0 = 35 + ( - 1.12t )

➔ 0 - 35 = - 1.12t

➔ - 35 = - 1.12t

➔ - 1.12t = - 35

➔ 1.12t = 35

➔ t = 35/1.12

➔ t = 3500/112

➔ t = 31.25 seconds

Hence,

Time taken by the train ( t ) = 31.25 seconds

Additional Information

Question

What relations does the equation of motion give ?

Answer

The 1st equation of motion gives us the relationship between the Final velocity and the time taken .

The 2nd equation of motion gives us the relationship between the displacement and the time taken .

The 3rd equation of motion gives us the relationship between the final velocity and the displacement.

By knowing this relationship enables us to know where to use the particular equation.

Question

What is retardation ? How to can we identify retardation without performing calculations ?

Answer

Negative acceleration is termed as Retardation .

SI unit of acceleration is m/s² .

If the initial velocity is greater than final velocity then it is said to be that the body is decelerating .

Question

Who was the scientist who did a lot of experiments and studies to understand motion ?

Answer

Galileo was the scientist who did a lot of experiments and studies to understand motion .

After , him Newton started his experiments and studies to study in more detail about motion .

But, firstly Archimedes was the one who thought about the word motion ! .

Answer 2

$\mathtt{\huge{\underline{\red{Question\:?}}}}$

✴ 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.

$\mathtt{\huge{\underline{\green{Answer:-}}}}$

✒ The acceleration is - 1.12 ms-² and the time taken to come to rest from the 35m/s is 31.3 seconds .

$\mathtt{\huge{\underline{\purple{Solution:-}}}}$

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.

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