Question

A particle moves along the x-axis so that at time t its position is given by s(t) = (t + 1)(t - 3)3, t > 0. /p>For what values of t is the velocity of the particle increasing? (2 points)
A.) t > 3
B. )1 < t < 3
C.) 0 < t < 1 and t > 3
D.) t > 0

Answer 1

Given the position function, the velocity function is obtained by taking the derivative:

$s(t)=(t+1)(t-3)^3\implies v(t)=(t-3)^3+3(t+1)(t-3)^2$

The velocity is increasing its own derivative is positive, so we also have to find the acceleration by taking another derivative:

$a(t)=4(t-3)^3+3(t-3)^2+6(t+1)(t-3)$

To find when $a(t)>0$, we first need to know where $a(t)=0$:

$4(t-3)^3+3(t-3)^2+6(t+1)(t-3)=(t-3)\bigg(4(t-3)^2+3(t-3)+6(t+1)\bigg)=(t-3)(4t^2-15t+33)=0$

The quadratic factor is always positive (its discriminant is negative), which leaves one solution at $t=3$. To either side of $t=3$, we have, for instance,

$a(2)=-12$

$a(4)=36>0$

which indicates that $v(t)$ is increasing for $t>3$, making the answer A.