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

Click For Summary
SUMMARY

The discussion focuses on the mechanics of a bullet fired from a rifle barrel, specifically analyzing the velocity function v(t) = (−5.00 ✕ 10^7)t² + (2.80 ✕ 10^5)t. The acceleration of the bullet is calculated using the derivative of the velocity function, resulting in a(t) = 2(5*10^5)t + 2.8*10^5, with the acceleration at the moment the bullet exits the barrel being 2.8*10^5 m/s². The participants also address the need to verify the derivative calculations and explore the implications for the bullet's acceleration duration and exit speed.

PREREQUISITES
  • Understanding of calculus, specifically differentiation
  • Familiarity with kinematic equations in physics
  • Knowledge of units of measurement in physics (m/s, m/s²)
  • Basic understanding of projectile motion
NEXT STEPS
  • Study the principles of differentiation in calculus
  • Learn about kinematic equations and their applications in projectile motion
  • Explore the concept of acceleration and its calculation from velocity functions
  • Investigate the physics of firearms and bullet dynamics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in the mathematical modeling of projectile motion.

EishaR
Messages
1
Reaction score
0
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
 
Physics news on Phys.org
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?
 

Similar threads

Replies
2
Views
1K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
648
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
14
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
9
Views
3K