University:
Angelo State University
Course:
ENGR 2302 | Engineering Mechanics - Dynamics
Academic year:

2024
Views:

355

Pages:

7
Author:

camilobonilladz36

Problem 12-14 A particle moves along a straight line such that its position is defined by (s = bt^2 + ct + d). Determine the average velocity, the average speed, and the acceleration of the particle when (t = t_1). 1 Given: (b = 1 \frac{m}{s^2}), (c = -6 \frac{m}{s}), (d = 5 m), (t_0 = 0 s), (t_1 = 6 s) Solution: (s_p(t) = bt^2 + ct + d) (v(t) = \frac{d}{dt} s_p(t)) (a(t) = \frac{d}{dt} v(t)) Find the critical time (t_2 = 2s) Given (v(t_2) = 0) (t_2 = 3 s) (v_{ave} = \frac{s_p(t_1) - s_p(t_0)}{t_1}) (v_{ave} = 0 \frac{m}{s}) (v_{speed} = \frac{|s_p(t_1) - s_p(t_2)| + |s_p(t_2) - s_p(t_0)|}{t_1}) (v_{speed} = 3 \frac{m}{s}) (a_I = a(t_1)) (a_I = 2 \frac{m}{s^2}) Problem 12-28 The acceleration of a particle along a straight line is defined by (a_p = bt + c). At (t = 0), (s_p = s_{p0}) and (v_p = v_{p0}). When (t = t_1), determine (a) the particle's position, (b) the total distance traveled, and (c) the velocity. Given: (b = 2 \frac{m}{s^3}), (c = -9 \frac{m}{s^2}), (s_{p0} = 1 m), (v_{p0} = 10 \frac{m}{s}), (t_1 = 9 s) Solution: (a_p = bt + c) (v_p = (\frac{b}{2})t^2 + ct + v_{p0}) (s_p = (\frac{b}{6})t^3 + (\frac{c}{2})t^2 + v_{p0}t + s_{p0}) a) The position (s_{p1} = (\frac{b}{6})t_1^3 + (\frac{c}{2})t_1^2 + v_{p0}t_1 + s_{p0}) (s_{p1} = -30.5 m) b) The total distance traveled - find the turning times (v_p = (\frac{b}{2})t^2 + ct + v_{p0} = 0) (t_2 = \frac{-c - \sqrt{c^2 - 2bv_{p0}}}{b}) (t_2 = 1.298 s) (t_3 = \frac{-c + \sqrt{c^2 - 2bv_{p0}}}{b}) (t_3 = 7.702 s) (s_{p2} = (\frac{b}{6})t_2^3 + (\frac{c}{2})t_2^2 + v_{p0}t_2 + s_{p0}) (s_{p2} = 7.127 m) (s_{p3} = (\frac{b}{6})t_3^3 + (\frac{c}{2})t_3^2 + v_{p0}t_3 + s_{p0}) (s_{p3} = -36.627 m) (d = |s_{p2} - s_{p0}| + |s_{p2} - s_{p3}| + |s_{p1} - s_{p3}|) (d = 56.009 m) c) The velocity (v_{p1} = (\frac{b}{2})t_1^2 + ct_1 + v_{p0}) (v_{p1} = 10 \frac{m}{s})