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 = (-5.35 107) t 2 + (3.05 105) 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.
(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.)
a = m/s2
x = m

(b) Determine the length of time the bullet is accelerated.
s

(c) Find the speed at which the bullet leaves the barrel.
m/s

(d) What is the length of the barrel?
m

Explanation / Answer

Given velocity    v =(-5.35 x107) t 2 +(3.05 x 105) t Accleration a = dv / dt                      =  2(-5.35 x107) t+ (3.05 x 105)                    =  (-10.7 x107) t + (3.05 x 105) Displacement or position    x =integral v dt                                             =  [(-5.35 x107) t 3/ 3 ] + (3.05 x 105) [ t 2 / 2]                                             = [(-1.783 x107) t 3 ] + (1.525 x 105) [ t 2 ] ----------------(1) Accleration = 0 at    end of thebarrel          (-10.7x107) t +(3.05 x 105) = 0                           t = 2.85 * 10 ^ -3 s Position at time t = 2.85 * 10 ^ -3 s  is   -----------------(2) substitute (2) in(1) x =  [(-1.783 x107) (  2.85 * 10 ^ -3)  3 ] + (1.525 x 105) ( 2.85 * 10 ^-3) 2 ]          = 1.6514 m So. length of the barallel L = 1.6514 m Given velocity    v =(-5.35 x107) t 2 +(3.05 x 105) t Accleration a = dv / dt                      =  2(-5.35 x107) t+ (3.05 x 105)                    =  (-10.7 x107) t + (3.05 x 105) Displacement or position    x =integral v dt                                             =  [(-5.35 x107) t 3/ 3 ] + (3.05 x 105) [ t 2 / 2]                                             = [(-1.783 x107) t 3 ] + (1.525 x 105) [ t 2 ] ----------------(1) Accleration = 0 at    end of thebarrel          (-10.7x107) t +(3.05 x 105) = 0                           t = 2.85 * 10 ^ -3 s Position at time t = 2.85 * 10 ^ -3 s  is   -----------------(2) substitute (2) in(1) x =  [(-1.783 x107) (  2.85 * 10 ^ -3)  3 ] + (1.525 x 105) ( 2.85 * 10 ^-3) 2 ]          = 1.6514 m So. length of the barallel L = 1.6514 m Accleration = 0 at    end of thebarrel          (-10.7x107) t +(3.05 x 105) = 0                           t = 2.85 * 10 ^ -3 s Position at time t = 2.85 * 10 ^ -3 s  is   -----------------(2) substitute (2) in(1) x =  [(-1.783 x107) (  2.85 * 10 ^ -3)  3 ] + (1.525 x 105) ( 2.85 * 10 ^-3) 2 ]          = 1.6514 m So. length of the barallel L = 1.6514 m

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