Question

A particle moves in a straight line with its position given by the following equation:x(t)=t^4-4t^3+2t^2+3t+6

a.Find its position after 1 second

b.Find its velocity after 2 seconds

Answer 1

Given:

$x(t)=t^4-4t^3+2t^2+3t+6$

(a) the position at the time $t=1\:\rm s$

$$x(1)=1^4-4*1^3+2*1^2+3*1+6=8\:\rm m$$

(b) the position velocity of the particle

$$v(t)=x'(t)=(t^4-4t^3+2t^2+3t+6)'\\ =4t^3-12t^2+4t+3$$

at the time $t=2\:\rm s$

$$v(2)=4*2^3-12*2^2+4*2+3=-5\:\rm m/s$$