A bullet leaves a rifle with a muzzle velocity of 521m/s . While accel

1.	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.
Answer» 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 $\sf{\bf{1.62\times\:10^{5} \: ms^{-2}}}$

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.

Answer»

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 $\sf{\bf{1.62\times\:10^{5} \: ms^{-2}}}$