Question

The speed of a train changes from 36 km/h to
72 km/h in 10 sec. Calculate the distance
travelled during this time.​

Answer 1

Required Solution:

Given:

• Initial velocity of the train (u) = 36km/h

→ $\sf { \cancel{36} \times \dfrac{5}{\cancel{18}} }$

→ 2 × 5

→ 10m/s

• Final velocity of the train (v) = 72km/h

→ $\sf { \cancel{72} \times \dfrac{5}{\cancel{18}} }$

→ 4 × 5

→ 20m/s

• Time taken (t) = 10s

To calculate:

• Distance travelled (s)

Calculation:

We know that,

• ${\underline {\boxed {\Large {\bf \gray { {v}^{2} - {u}^{2} = 2as } }}}}$

Let's calculate the acceleration first:

By using the first equation of motion:

• v = u + at

→ 20 = 10 + ( a × 10 )

→ 20 = 10 + 10a

→ 10a = 20 - 10

→ 10a = 10

→ a = 10/10

→ a = 1m/s^2

Now,

• v² – u² = 2as

→ (20)² - (10)² = 2 × 1 × s

→ 400 - 100 = 2s

→ 300 = 2s

→ s = $\cancel{\dfrac{300}{2}}$

→ s = 150 m [ Distance]

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More information!

Equations of motion:

• v = u + at
• s = ut + ½at²
• v² – u² = 2as

Where,

• ★ v = final velocity = m/s
• ★ u = initial velocity = m/s
• ★ a = acceleration = m/s²
• ★ s = distance/displacement = m
• ★ t = time = sec

Remember that!

• When a body starts from rest, its initial velocity is 0.
• When a body comes to stop or applies breaks, its final velocity is 0.

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Answer 2

Intitial velocity = 36 kmph

Final velocity = 72 kmph

Time = 10 s

We know,

a = (v - u) × t^-1

⇒a = (72 km/h - 36 km/h)/(10 s)

⇒a = (36 km/h)/(10 s)

{Multiply 5/18 with the speed given in km/h units.}

⇒a = (10 m/s)/(10 s)

⇒a = 1 m/s²

Now, we know,

s = ut + ½at²

⇒s = (10 m/s)(10 s) + ½(1 m/s²)(10 s)²

{Convert in a similar manner}

⇒s = 100 m + ½(100 m)

⇒s = 150 m {Solution}