Answer:
The acceleration of the object is -69.78 m/s²
Explanation:
Given;
postion of the particle:
[tex]x = 32t - 38\frac{t^3}{3} \\\\[/tex]
The velocity of the particle is calculated as the change in the position of the particle with time;
[tex]v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s[/tex]
Acceleration is the change in velocity with time;
[tex]a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2[/tex]
