The speed of a bullet as it travels down the barrel of a rifle toward the openin

ID: 1334209 • Letter: T

Question

The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by

V = -5*107 t2 + 3*105 t, where v is in meters per second (m/s).

The acceleration of the bullet as just leaves the barrel is zero.

Show your motion diagram and complete solution.

Determine the position and the acceleration of the bullet as a function of time when the bullet is in the barrel.

Determine the time interval over which the bullet is accelerated.

Find the speed at which the bullet leaves the barrel

d) What is the length of the barrel?

given,

V = -5*107 t2 + 3*105 t

acceleration = dV / dt

acceleration = d(-5*107 t2 + 3*105 t)/dt

acceleration = 300000-100000000 t

V = ds / dt

ds = (-5*107 t2 + 3*105 t)dt

position s = 150000 t^2-(50000000 t^3)/3

300000-100000000 t = 0

t = 0.003 sec

bullet is accelerated for 0.003 sec

speed after 0.003 sec

V = -5 * 10^7 * 0.003^2 + 3 * 10^5 * 0.003

speed at which bullet leaves the barrel = 450 m/s

distance covered in 0.003 sec

distance = 150000 * 0.003^2 - (50000000 * 0.003^3)/3

length of the barrel = 0.9 m

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.