A bullet leaves a rifle with a muzzle velocity of 1042 m/s. while accelerating through the barrel of the rifle, the bullet moves a distance of 1.680 m. determine the acceleration of the bullet (assume a uniform acceleration)
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Answered by
51
The question states that 0.840 m is the distance travelled within the barrel. Therefore u = 0 m/s, v= 521 m/s.
This means that a = (v^2 - u^2)/2s.
This is a reformat of v^2 = u^2 + 2as
a = (521^2)/(2 x 0.840)
a = 161.57 m/s^2
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This means that a = (v^2 - u^2)/2s.
This is a reformat of v^2 = u^2 + 2as
a = (521^2)/(2 x 0.840)
a = 161.57 m/s^2
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Answered by
82
Answer:-
= 323144.04 m/s²
Given
u = 0
v = 1042 m/s
s = 1.680 m
Using third equation of motion:-v²-u² = 2as
⇒v²-u² = 2as
⇒1042 m/s² - 0 = 2a × 1.680
⇒ a = 1042 × 1042/2 × 1.680
⇒ a = 1085764/3.36
⇒ a = 323144.04 m/s²
Acceleration of the bullet is 323144.04 m/s²
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