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 = (–4.3 × 10^7)t^2 + (2.00 × 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 a function of time when the bullet is in the barrel. a= and x=
(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?

Explanation / Answer

a) a = dv / dt = (-8.60 * 107) t + (2.00 * 105 )

dx / dt = v , dx = v * dt

x = (-1.43 * 107) t3 + (1.00 * 105 ) t2 + C [c = 0]

x = (-1.43 * 107) t3 + (1.00 * 105 ) t2

b) a = 0 = (-8.60 * 107) t + (2.00 * 105 )

Length of time the bullet is accelerated t = 2.33 * 10-3 s

c) v = (-4.3 * 107) t2 + (2.00 * 105)t [ t = 2.33 x 10-3 s ]

v = (-4.3 * 107) (2.33 x 10-3)2 + (2.00 * 105)(2.33 x 10-3)

speed at which the bullet leaves the barrel v = 233 m/s

d) length of the barrel

x = (-1.43 * 107) t3 + (1.00 * 105 ) t2 [ t = 2.33 x 10-3 s]

x = 0.362 m

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