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Need help with Kinematics problem?
A gun accelerates a bullet at rest with an acceleration of 6000 m/s^2. If the muzzle of the gun is 20 cm long...
a) How fast is the bullet traveling when it leaves the gun?
b) How long does it take the bullet to exit?
I'm so confused. There's like not enough information to solve it, I think. I need time to find Velocity/Speed. I need Velocity/Speed to solve for time.
2 Answers
- FiremanLv 79 years agoFavorite Answer
(a) By v^2 = u^2 + 2as
=>v^2 = 0 + 2 x 6000 x 20 x 10^-2
=>v = √2400
=>v = 48.99 m/s
(b) By v = u + at
=>48.99 = 0 + 6000 x t
=>t = 8.16 x 10^-3 sec or 8.16 ms
- PearlsawmeLv 79 years ago
A)
v² = u² + 2 as = 0² - 2*6000*20e-2
v = 48.99 m/s
B)
Distance = average velocity * time
average velocity = 48.99/2 =24.49 m/s
Time = 20e-2 / 24.49 =0.008 s
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