Answer to Question #271797 in Mechanics | Relativity for Happpy

Question

1. The position of a particle which moves along the straight line is define by the relation 𝑥(𝑡) = 𝑡^3 − 6𝑡^2 − 15𝑡 + 40 where 𝑥 and 𝑡 are expressed in meters and seconds respectively. Note that the coefficients of t have dimensions accordingly. (a) Determine when the velocity of the particle is zero. (b) Calculate the position vector and distance travelled by the particle when the acceleration is zero. Consider that at the starting point time 𝑡 = 0 sec. (c) Does the particle move at constant velocity or constant acceleration? Justify your answer

Answer 1

(a) Determine when the velocity of the particle is zero:

$$v(t)=x'(t)=3t^2-12t-15=0,\\ t=5\text{ s when } v(t)=0.$$

(b) Calculate the position vector and distance travelled by the particle when the acceleration is zero.

$$a(t)=v'(t)=6t-12=0,\\ t=2\text{ s when }a(t)=0.\\\space\\ x(2)=2^3-6·2^2-15·2+40=-6\text{ m},\\\space\\ d(2)=\int^2_0v(t)\text dt=-46\text{ m}.$$

(c) The particle moves at varying velocity and increasing acceleration because the acceleration is directly proportional to time, while velocity is proportional to time squared.