Question

A bullet leaves a rifle with a muzzle velocity of 521m/s . While acceleration 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.
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Answer 1

Given :

• Muzzle velocity of the bullet = 521 m/s

• Distance covered by the bullet = 0.840 m

To find :

• The acceleration of the bullet

Solution :

Assume that the initial velocity of the bullet as zero (as it leaves the barrel)

Now , we have

• v = 521 m/s
• u = 0 m/s
• s = 0.840 m

Applying second equation of motion ,

$\\ {\boxed{ \bf{ {v}^{2} - u {}^{2} = 2as}}}$

Here ,

• v is final velocity
• u is initial velocity
• a is acceleration
• s is distance covered

Substituting the values ,

$\\ : \implies \bf \: (521 \: m {s}^{ - 1} ) {}^{2} - (0) {}^{2} = 2(a)(0.840 \: m) \\ \\ \\ : \implies \bf \: (521 \times 521) \: m {}^{2} {s}^{ - 2} - 0 = 1.68 \: m(a) \\ \\ \\ : \implies \bf \: 271441 \: m {}^{2} {s}^{ - 2} = 1.68 \: m(a) \\ \\ \\ : \implies \bf \: \frac{271441 \: m {}^{2} {s}^{ - 2} }{1.68 \: m} = a\\ \\ \\ : \implies \bf \: 161572.02 \: m {s}^{ - 2} = a \\ \\ \\ : \implies \bf \: a \approx \: 162000 \: m {s}^{ - 2} \\ \\ \\ : \implies \bf \: a \approx \: 162 \times {10}^{4} \: m {s}^{ - 2} \\ \\ \\ : \implies {\boxed {\underline{\bf \: a \approx \: 1.62 \times {10}^{5} \: m {s}^{ - 2} }}}$

Hence , The Acceleration of the bullet is $\sf{\bf{1.62\times\:10^{5} \: ms^{-2}}}$

Answer 2

$\huge \bf AnSwEr :$

final velocity of the bullet is 521 m/s

Distance covered by the bullet is 0.840 m

To find, accerlation of the bullet

Solution :-

by using 3rd equation of motion

➝ v² - u² = 2as

➝ (521)² - (0)² = (2)(a)(0.840)

➝ 271441 = 1.68a

➝ a = 271441/1.68

➝ a = 161572.02 m/s²