Question

A particle moves along a straight line. Its position at any instant is given by x = 32t− 38t^3/3 where x is in metre and t in second. Find the acceleration of the particle at the instant when particle is at rest.

Answer 1

Answer:

The acceleration of the object is -69.78 m/s²

Explanation:

Given;

postion of the particle:

$x = 32t - 38\frac{t^3}{3}$

The velocity of the particle is calculated as the change in the position of the  particle with time;

$v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s$

Acceleration is the change in velocity with time;

$a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2$