Dynamics Chapter 1 - Question 1.8 | Georgia Southern University - Edubirdie

University:
Georgia Southern University
Course:
ENGR 2232 | Dynamics of Rigid Bodies
Academic year:

2024
Views:

492

Pages:

1
Author:

Jalen W.

Problem 1.8: The acceleration of a point P, moving along a straight line, is directed towards the point Z, and its magnitude is inversely proportional to the distance x . For t = 0 , the point P has the distance x_0 = 2 m, the velocity v_0 = 4 m/s, and the acceleration a_0 = -3 m/s^2. a) Determine the velocity v_1 for the distance x_1 = 3 m. b) At what distance x_2 is the velocity zero? Matlab Code: % Given data x0 = 2; % initial distance in meters v0 = 4; % initial velocity in m/s a0 = -3; % initial acceleration in m/s^2 x1 = 3; % distance for part a) in meters % a) Determine the velocity v1 for the distance x1 % Since acceleration is inversely proportional to distance, we have a = k/x % Integrating this expression gives v = k * ln(x) + C, where C is the constant of integration % At t = 0, x = x0 and v = v0 % So, v0 = k * ln(x0) + C % Similarly, a0 = k / x0 % Solving these equations gives k and C k = a0 * x0; C = v0 - k * log(x0); % Using the derived expression for velocity, calculate v1 v1 = k * log(x1) + C; % b) At what distance x2 the velocity is zero? % To find when velocity is zero, solve v = 0 = k * ln(x) + C for x % x = exp(-C/k) x2 = exp(-C/k); % Display results fprintf('Velocity v1 for x1 = 3m: %.2f m/s\n', v1); fprintf('Distance x2 where velocity is zero: %.2f m\n', x2);

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