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

Question

The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v =(-4.30*10^7)t^2+(2.25*10^5)t. The acceleration of the bullet as it leaves the barrel is zero.
(a) Determine the acceleration and position of the bullet as a function of time when the bullet is in the barrel. use t as necessary and round all numerical coefficients to exactly 3 significant figures.
(b) Determine the length of time the bullet is accelerated.
(c) Find the speed at which the bullet leaves the barrel. (d) What is the length of the barrel?
(d) What is the length of the barrel?

Any help would be appreciated.Thank you.

Explanation / Answer

(a) a = dv/dt = (-4.3*10^7)*2t + 2.25*10^5 = -8.6*10^7t + 2.25*10^5

(b) because at the end of tbe barrel, a = 0

0 = -8.60*10^7 t + 2.25*10^5

t = 2.25*10^5 /(8.6*10^7) = 2.62*10^-3 s

(c) plug t = 2.62*10^-3 into equation for v

v = (-4.30*10^7)(2.62*10^-3)^2 + (2.25*10^5)(2.62*10^-3)

= -2.95*10^2 + 5.895*10^2 = 2.945 * 10^2 m/s

(d) L = intigration vdt = (-4.30*10^7)(1/3)t^3+(2.25*10^5)*(1/2)t^2

plug in t = 2.62*10^-3 s

L = (-4.30*10^7)(1/3)(2.62*10^-3)^3 + (2.25*10^5)(1/2)(2.62*10^-3)^2

= -2.58*10^-1 + 7.72*10^-1 = 5.14*10^-1 m

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