12. The speed of a bullet as it travels down the barrel of a rifle toward the op

Question

12. The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v -5.00712 3.00 x 105t Where v is in meters per square second and t is in seconds. The acceleration of the bullet just as it leaves the barrel is zero Determine the acceleration and position of the bullet as functions of time when the bullet is in the barrel Determine the time interval over which the bullet is accelerated Find the speed at which the bullet leaves the barrel What is the length of the barrel? Can we use the kinematics equation to analyze this problem? Why or Why not?

Explanation / Answer

a) acceleration; a = dv/dt = (-10.0 x 10^7)t + 3.0 x 10^5
dx/dt = v
dx = v dt
position; x = (-1.67 x 10^7)t^3 + (1.50 x 10^5)t^2 + C

x = 0 t = 0 C = 0
x = (-1.67 x 10^7)t^3 + (1.50 x10^5)t^2

b) a = 0 = (-10.0 x 10^7)t + (3.0 x 10^5)
time interval t = 3.0 x 10^-3 sec

c)v = (-5.0 x 10^7)t^2 + (3.0 x 10^5)t @ t=3.0 x 10^-3 sec
v = 450 m/s

d) the length of the barrel x = (-1.67 x 10^7)t^3 + (1.50 x10^5)t^2 @ t = 3.0 x 10^-3 sec.
x = -0.4509 +1.35 = 0.8991 m

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