Question

A bullet leaves a rifle with a muzzle velocity of 521
m/s. While accelerating through the barrel of the
rifle, the bullet moves a distance of 0.840 m.
Determine the acceleration of the bullet (assume a
uniform acceleration).​

Answer 1

Step-by-step explanation:

I hope you have understood the question

Answer 2

Given :

• A bullet leaves a rifle with a muzzle velocity =521m/s

• The bullet moves a distance =0.840m

To Find :

• The acceleration of the bullet =?

Formula :

$\\ \red \bigstar \boxed{ \bf{ {v}^{2} - {u}^{2} = 2ad}}$

Solution :

• We assume the bullet leaves from rest

So, the initial velocity will be ,u=0 m/s

•Also given, final velocity (v)=512 m/s

•Distance travelled (d) =0.840m

Let a be the acceleration of the bullet.

$\\ \star\large\pink{\underline{ \sf{By \: using \: formula}}}$

$\\ \\ \implies \sf \: {v}^{2} - {u}^{2} = 2ad \\ \\ \\ \implies \sf \: {(512)}^{2} - {(0)}^{2} = 2a(0.840) \\ \\ \\ \implies \sf \: {(512)}^{2} = 2a(0.840) \\ \\ \\ \implies \sf \: \frac{512 \times 512}{0.840} = 2a \\ \\ \\ \implies \sf \: a = \frac{256 \times 512}{0.840} \\ \\ \\ \implies \sf \: a = 1.62 \times {10}^{5} \\ \\ \\ \implies \boxed{ \sf{ \therefore \: a = 1.62 \times {10}^{5} ms^{-2}}}$

The acceleration of the bullet will be

$\blue\star\underline{\sf{a=1.62 \times 10^{5}ms^{-2}}}$