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 = (5.85 107)t2 + (2.95 105)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 = c m/s2 x = c 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?

Explanation / Answer

a)

v = (5.85 x 10^7)t2 + (2.95 x 10^5)t

a = dv/dt = (-11.7 x 10^7)t + ( 2.95 x 10^5)

dx/dt = v

dx = v*dt

Integrating w.r.t 't'

x = (-1.95 x 10^7)t^3 + (1.475 x 10^5)t^2 + C

x = 0 t = 0 C = 0

x = (-1.95 x 10^7)t^3 + (1.475 x10^5)t^2

b)

a = 0 = (-11.7 x 10^7)t + (2.95 x 10^5)

t = 2.52 x 10^-3 sec

c)

v = (-5.85 x 10^7)t^2 + (2.95 x 10^5)t

t = 2.52 x 10^-3 sec

v = 371.9 m/s

d)

x = (-1.95 x 10^7)t^3 + (1.475 x10^5)t^2

t = 2.52 x 10^-3 sec.

x = 0.625 m

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