Question

A particle moves along the x-axis according to the equation: x=3.0 t3 + 4.0 t4 -2.0 t where x is in meters and t is in seconds. At t=3 seconds, find: a. the position of the particle b. the velocity at that instance c. the acceleration at that instance

Answer 1

The equation of motion of a particle

$$x(t)=3t^3+4t^4-2t$$

The velocity

$$v(t)=x'(t)=9t^2+16t^3-2$$

The acceleration

$$a(t)=v'(t)=18t+48t^2$$

Hence,

$$x(3)=318\:{\rm m},\quad v(3)=511\:{\rm m/s}, \\ a(3)=486\:{\rm m/s^2}.$$