Question

train moving with velocity 20m/s is brought to rest in 5

second by applying break find the distance coverd by it.​

Answer 1

Given :

• Initial velocity, u = 20 m/s
• Final velocity, v = 0 m/s
• Time, t = 5 seconds

To find :

• Distance covered

According to the question,

By using Newtons first equation of motion we will find Acceleration (a) then by using Newtons second equation of motion we will find distance covered.

→ v = u + at

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• t = Time

→ 0 = 20 + a × 5

→ 0 = 20 + 5a

→ 0 - 20 = 5a

→ -20 = 5a

→ -20 ÷ 5 = a

→ - 4 = a

So,the acceleration is - 4 m/s². Negative signs means retardation.

Now,

By using Newtons second equation of motion,

→ v² = u² + 2as

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• s = Distance

→ Substituting the values,

→ (0)² = (20)² + 2 × (-4) × s

→ 0 = 400 + (-8s)

→ 0 - 400 = - 8s

→ - 400 = - 8s

→ 400 ÷ 8 = s

→ 50 = s

So,the distance covered is 50 metres.

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More formulas :

Newton's first equation of motion :

• v = u + at

Newton's second equation of motion :

• s = ut + ½ at²

Newton's third equation of motion :

• v² = u² + 2as

Answer 2

Given -

• Initial velocity (u) = 20 m/s
• Final velocity (v) = 0

Time (t) = 5s

To find -

• Distance covered by the train.

Solution -

• v = u + at

Substitute the given values.

↬ 0 = 20 + a × 5

↬ -20 = a × 5

↬ $\sf{\dfrac{-20}{50}}$ = a

↬ - 4 = a

Retardation = 4 $\sf{m/s^2}$

• $\sf{v^2 - u^2 = 2as}$

Substitute the given values.

↬ $\sf{(0)^2 - (20)^2 = 2 \times - 4 \times s}$

↬ $\sf{-400 = - 8 \times s}$

↬ s = 50 m

$\overbrace{\underbrace{\sf{\green{Distance = 50~m}}}}$

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Know more -

Velocity -

• It is distance travelled by a body in one second in a particular direction.

• It is a vector quantity.

• Its S.I. unit is m/s.

Acceleration -

• It is the rate of change of velocity of a body.

• It is vector quantity.

• Its S.I. unit is $\sf{m/s^2}$

• Retardation is the negative acceleration.

Equations of motion -

• First equation of motion -

↬v = u + at

• Second equation of motion -

↬s = ut + $\sf{\dfrac{1}{2} at^2}$

• Third equation of motion -

↬$\sf{v^2 - u^2 = 2as}$

where,

v = Finial velocity

u = Initial velocity

a = acceleration

s = distance/displacement

t = time