Question

The position of a point is given as a function of time by x= 4t − 6t + 2t− 1 , where x and t are expressed in meters and seconds. Determine the position, the velocity, and the acceleration of the particle when t= 2 s.

Answer 1

Given:

$x=4-6t+2t^2$

$t=2\:\rm s$

The position

$$x(2)=4-6*2+2*2^2=0$$

The velocity

$$v(2)=x'(2)=-6+4*2=2\:\rm m/s$$

The acceleration

$$a(2)=v'(2)=4\:\rm m/s^2$$