Physics 101 mechanics help -- Bullet fired by a rifle barrel

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The discussion focuses on calculating the acceleration and position of a bullet fired from a rifle, with the speed equation given as v = (−5.00 ✕ 10^7)t² + (2.80 ✕ 10^5)t. The acceleration is derived as a = dv/dt, resulting in a = 2(5*10^5)t + 2.8*10^5, indicating the bullet's acceleration is 2.8*10^5 m/s² at t=0. There is confusion regarding the sign change in the derivative, prompting a request for clarification on the calculations. The thread emphasizes the importance of checking derivatives and completing all parts of the problem for accurate results.
EishaR
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Homework Statement
A repeating rifle fires 2 LR bullets such that as they travel down the barrel of the rifle their speed is given by v = (−5.00 ✕ 107)t2 + (2.80 ✕ 105)t,where v is in meters per second and t is in seconds. The acceleration of the bullet just as it leaves the barrel is zero.
Determine the acceleration (in m/s2) and position (in m) of the bullet as a function of time when the bullet is in the barrel. (Use the following as necessary: t. Round all numerical coefficients to at least three significant figures. Do not include units in your answers. Assume that the position of the bullet at t = 0 is zero.)
a (t)=
x(t)=

(b)Determine the length of time the bullet is accelerated (in s).

(c) Find the speed at which the bullet leaves the barrel (in m/s).

(d) What is the length of the barrel (in m)?
Relevant Equations
y v = (−5.00 ✕ 10^7)t2 + (2.80 ✕ 10^5)t,
a) a= dv/dt = 2(5*10^5)t+2.8*10^5
2(5*10^5)(0)+2.8*10^5=2.8*10^5m/s^2
 
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EishaR said:
Homework Statement:: A repeating rifle fires 2 LR bullets such that as they travel down the barrel of the rifle their speed is given by v = (−5.00 ✕ 107)t2 + (2.80 ✕ 105)t,where v is in meters per second and t is in seconds. The acceleration of the bullet just as it leaves the barrel is zero.
Determine the acceleration (in m/s2) and position (in m) of the bullet as a function of time when the bullet is in the barrel. (Use the following as necessary: t. Round all numerical coefficients to at least three significant figures. Do not include units in your answers. Assume that the position of the bullet at t = 0 is zero.)
a (t)=
x(t)=

(b)Determine the length of time the bullet is accelerated (in s).

(c) Find the speed at which the bullet leaves the barrel (in m/s).

(d) What is the length of the barrel (in m)?
Relevant Equations:: y v = (−5.00 ✕ 10^7)t2 + (2.80 ✕ 10^5)t,

a) a= dv/dt = 2(5*10^5)t+2.8*10^5
2(5*10^5)(0)+2.8*10^5=2.8*10^5m/s^2
For the first part, check your derivative. You must attempt the other parts also.
 
Last edited:
How did −5.00 ✕ 10^7 turn into +5.00 ✕ 10^5?
 
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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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