# physics

The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by
v=(-5.00×10^7)t^2 + (3.00×10^5)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. (a) Determine the acceleration and position of the bullet as functions of time when the bullet is in the barrel. (b) Determine the time inter- val over which the bullet is accelerated. (c) Find the speed at which the bullet leaves the barrel. (d) What is the length of the barrel?

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1. I will do position last

a = dv/dt
so
a = -10*10^7 t + 3*10^5

zero at
t = 3/10 * 10^-2 = .003 seconds
a is positive for the whole .003s

find v at t = .003 for part c

x = integral v dt
with x = 0 at t = 0
so
x = (1/3)(-5.00×10^7)t^3 +(1/2)(3.00×10^5)t^2
length of barrel is x when t = .003

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posted by Damon

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