Question

A bullet leaves a barrel of a rifle with speed of 300 m/sec if the length of the barrel is 0.9m at what rate the bullet acceleration while in barrel.

Answer 1

Answer : Acceleration of the bullet is $50000\ m/s^2$.

Explanation:

Given that,

Speed of the bullet, v = 300 m/s

Length of the barrel, d = 0.9 m

We have to find the acceleration of the barrel. Using third equation of motion as :

$v^2-u^2=2as$

$a=\dfrac{v^2-u^2}{2s}$

a is the acceleration of the barrel.

$a=\dfrac{(300\ m/s)^2-0}{2\times 0.9\ m}$

$a=50000\ m/s^2$

Hence, this is the required solution.

Answer 2

Required Solution:

Given:

• Initial velocity of the bullet (u) = 300m/s

• Length of the barrel (s) = 0.9m

• Final velocity of the bullet (v) = 0

To calculate:

• Acceleration of the bullet (a)

Calculation:

By substituting the values of all the given quantities in the 3rd equation of motion:

• 2as = v² – u²

→ 2 × a × 0.9 = (0)² – (300)²

→ 1.8a = 0 - 90,000

→ 1.8a = -90,000

→ a = $\sf { \dfrac{-90,000}{1.8} }$

→ a = $\sf { \dfrac{-90000}{18} \times 10 }$

→ a = $\sf { \dfrac{-90,000 \times 10}{18} }$

→ a = $\cancel{\sf { \dfrac{-9,00,000}{18} } }$

→ a = -50,000 m/s² [ Acceleration ]

Therefore, acceleration of the bullet is -50,000 m/s².

Let's verify it also:–

• 2as = v² – u²

→ 2 × (-50000) × 0.9 = 0² - 300²

→ -100,000 × 0.9 = -300²

→ -90,000 = -90,000

LHS = RHS

Hence, verified!

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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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