Question

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The speed of a bullet as it travels down the barrel of a rifle toward the open end is given by: v-(-5.00 x 107)t2 + (3.00 x 1 jt where v is in m/s and t is in seconds. The acceleration of the bullet just as it leaves the end of the barrel is zero. a. Find the position x and acceleration a of the bullet as functions of time while it is still within the barrel of the rifle. b. Determine the time interval over which the bullet is accelerated c. Find the speed at which the bullet exits the end of the barrel. d. What is the length of the barrel?

Answer 1

Substituting t from (b) v = (- 5.00 times 10^7) 93 times 10^- 3)^2 + (3.00 times 10^5) (3 times 10^- 3) m/s x = l = (- 5.00 times 10^7) (3 times 10^- 3)^3/3 + (3.00 times 10^5) (3 times 10^- 3)^2/2 = - 0.45 + 1.35 = 0.9 M at end a = 0 (- 10 - 00 times 10^7) + (3.00 times 10^5) = 0 Therefore t = 3 times 10^- 3s v = (- 5.00 times 10^7) t^2 + (3.00 times 10^5) t a = dv/dt = (- 5.00 times 10^7) 2t + (3.00 times 10^5) a = (- 10.00 times 10^7) t + (3.00 times 10^5) v = dx/dt Therefore integral x dt = integral dx = x Therefore integral v dt = integral [(- 5.00 times 10^7) t^2 + (3.00 times 10^5) t]dt