Question

Motion of a Bullet in Water A bullet shot underwater will experience "viscous drag" and follow this velocity equation: where v 223 m/s and b 0.28 s 1. The bullet has an initial position x, 33.0 m at What is the location of the bullet at time t 0.71 seconds? 0s. 1pts Submit Answer Tries 0/15 What is the maximum position of the bullet (if you wait a long time, what will its position be)? 1pts Submit Answer Tries 0/15 What is the x component of the acceleration of the bullet 0.5 seconds after the bullet is fired? 1pts Submit Answer Tries 0/15

Answer 1

Equation doublerightarrow v_x (t) = v_i middot e^-bt *. v_i = 223 m/s b = 0.28 s^-1 x_i = 33m t_i = 0 s doublerightarrow v_i/b = 223/0.28 = 796.43 * v_x (t) = v_i e^-bt doublerightarrow dx/dt = v_i e^-bt doublerightarrow integral^x_33 dx = integral^t_0 v_i middot e^-bt dt doublerightarrow x - 33 = v_i [e^-bt/-b]^t_0 doublerightarrow x = 33 + v_i/b [-e^-bt + 1] doublerightarrow x = 33 + v_i/b - v_i/b middot e^-bt doublerightarrow x = 33 + 796.43 - 796.43 middot e^-0.28 t therefore x = 829.43 - 796.43 middot e^-0.28 t \r\n (a) * At t = 0.71 s * location x = 829.43 - 796.43 middot e^-0.28 t x = 829.43 - 796.43 middot e^-0.28 times 0.71 therefore x = 176.585 m (b) * Maximum position of bullet t = alpha doublerightarrow x/max = 829.43 m (c) * a_x = dv_1/dt = v_i (-b) e^-bt = -v_1 b e^-bt a_x Vertical-bar _t = 0.5 s = -v_i b e^-b times 0.5 = -223 times 0.28 times e^-0.28 times 0.5 therefore a_x = -54.2827 m/sec^2