Question

Class 9th, Physics.

The speed of a train increases at a constant rate of $\alpha$ from zero to v, and then remains constant for an interval, and finally decreases to zero at a constant rate of $\beta$. If $\rm L$ be the total distance described, prove that the total time taken is :

$\rm { \dfrac{L}{v} + \dfrac{v}{2}\Bigg (\dfrac{1}{\alpha} + \dfrac{1}{\beta} \Bigg ) }$​

Answer 1

Explanation :-

For 1st case :-

Let the distance travelled in this case be s₁ . Also, let the time taken to cover the assumed distance be t₁ .

By using, 1st equation of motion :-

⇒ v = u + at

⇒ v = 0 + α(t₁)

⇒ t₁ = v/α ---(1)

Now, by the 3rd equation :-

⇒ v² - u² = 2as

⇒ v² = 2(α)(s₁)

⇒ s₁ = v²/2α

For 2nd case :-

Let the distance travelled be s₂ and time be t₂ . The train was moving with constant velocity (v), so we have :-

⇒ Distance = Speed × Time

⇒ s₂ = vt₂

For 3rd case :-

Let the distance covered be s₃ and time be t₃ . Final velocity of the train will be 0, initial velocity will be "v" and acceleration will be (-β) as train is retarding .

⇒ v = u + at

⇒ 0 = v + (-β)(t₃)

⇒ t₃ = v/β ----(2)

⇒ v² - u² = 2as

⇒ 0 - v² = 2(-β)(s₃)

⇒ s₃ = v²/2β

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Total distance covered by the train is L , so from here we get :-

⇒ L = s₁ + s₂ + s₃

⇒ L = v²/2α + vt₂ + v²/2β

⇒ L = v²/2(1/α + 1/β) + vt₂

⇒ L/v = v/2(1/α + 1/β) + t₂

⇒ t₂ = L/v - v/2α - v/2β ---(3)

Now, we can get the required equation for the total time taken by the train by adding equations (1), (2) and (3) .

= t₁ + t₂ + t₃

= v/α + L/v - v/2α - v/2β + v/β

= L/v + (v/α - v/2α) + (v/β - v/2β)

= L/v + v/2α + v/2β

= L/v + v/2(1/α + 1/β)

Hence, proved !

Answer 2

Given information :-

The speed of a train increases at a constant rate of α from zero to v, and then remains constant for an interval, and finally decreases to zero at a constant rate of β. If L be the total distance described, prove that the total time taken is :

• L/v + v/2 (1/α + 1/β)

Formula used,

• 1st eqⁿ of motion (v = u + at)
• 2nd eqⁿ of motion (s = ut + 1/2 at²)
• Distance = Speed × Time

Proof :-

Let,

• Distance travelled in 1st case be s₁
• Time taken in 1st case t₁
• Distance travelled in 2nd case be s₂
• Time taken in 2nd case be t₂
• Distance travelled in 3rd case be s₃
• Time taken in 3rd case be t₃

Case – ❶

Putting all values in 1st eqⁿ of motion,

➻ v = 0 + α(t₁)

➻ v = α × t₁

➻ v/α = t₁

➻ t₁ = v/α

Putting all values in 2nd eqⁿ of motion,

➻ s₁ = 0(t₁) + 1/2 (α)(t₁)²

➻ s₁ = (0 × t₁) + 1/2 × α × (v/α)²

➻ s₁ = 0 + 1/2 × α × v²/α²

➻ s₁ = 1/2 × v²/α

➻ s₁ = (1 × v²)/(2 × α)

➻ s₁ = v²/2α

Case – ❷

Putting all values in formula of distance,

➻ s₂ = v × t₂

➻ s₂ = vt₂

Case – ❸

• Here, final velocity of train will be 0, initial velocity of train will be v and acceleration of train will be -β (Sign is negative because train is retarding).

Putting all values in 1st eqⁿ of motion,

➻ 0 = v + (-β)(t₃)

➻ 0 - v = -β × t₃

➻ -v/-β = t₃

➻ v/β = t₃

➻ t₃ = v/β

Putting all values in 2nd eqⁿ of motion,

➻ s₃ = v(t₃) + 1/2 (-β)(t₃)²

➻ s₃ = (v × v/β) + 1/2 × (-β) × (v/β)²

➻ s₃ = v²/β + 1/2 × (-β) × v²/β²

➻ s₃ = v²/β + 1/2 × (-1) × v²/β

➻ s₃ = v²/β + (1 × -1 × v²)/(2 × β)

➻ s₃ = v²/β + (-v²/2β)

➻ s₃ = v²/β - v²/2β

➻ s₃ = (2v² - v²)/2β

➻ s₃ = v²/2β

Now, finding total distance travelled (L),

➻ L = Sum of all distances

➻ L = s₁ + s₂ + s₃

➻ L = v²/2α + vt₂ + v²/2β

➻ L = v²/2(1/α + 1/β) + vt₂

➻ L = v[v/2(1/α + 1/β) + t₂]

➻ L/v = v/2(1/α + 1/β) + t₂

➻ L/v = v/2α + v/2β + t₂

➻ L/v - (v/2α + v/2β) = t₂

➻ L/v - v/2α - v/2β = t₂

➻ t₂ = L/v - v/2α - v/2β

Now, finding total time taken,

➻ t₁ + t₂ + t₃

➻ v/α + L/v - v/2α - v/2β + v/β

➻ L/v + (v/α - v/2α) + (v/β - v/2β)

➻ L/v + [(2v - v)/2α] + [(2v - v)/2β]

➻ L/v + v/2α + v/2β

➻ L/v + v/2(1/α + 1/β)

• Hence, Proved ✔

Additional information :-

Three equations of motion,

1. v = u + at
2. s = ut + ½ at²
3. v² = u² + 2as

Where,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time taken
• s denotes distance travelled

Some important definitions,

• Acceleration

Acceleration is the process where velocity changes. Since, velocity is the speed and it has some direction. So, change in velocity is considered as acceleration.

• Initial velocity

Initial velocity is the velocity of the object before the effect of acceleration.

• Final velocity

After the effect of acceleration, velocity of the object changes. The new velocity gained by the object is known as final velocity.

• Distance travelled

The total path length covered by an object is called distance travelled.

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