Answer to Question #334098 in Calculus for Kirby

Question

Suppose that a particle moves along a line with position function s(t) =2t^2 +3t+1 where s is in meters and t is in seconds.

a. What is its initial position?

b. Where is it located after t = 2 seconds?

c. At what time is the particle at position s = 6?

Answer 1

a.

$s(0)=s_0=2(0)^2 +3(0)+1=1(m)$

b.

$s(2)=2(2)^2 +3(2)+1=15(m)$

15 meters from its initial position.

c.

$s(t)=6 m, t\ge0$

$2t^2 +3t+1=6$

$2t^2 +3t-5=0$

$(t-1)(2t+5)=0$

$t=1\ or\ t=-2.5$

Since $t\ge 0,$ we take $t=1 s.$

The particle will be at position s = 6 after 1 second.