# Thread: Velocity and Rate of Change

1. ## Velocity and Rate of Change

How do I find the values for t if the velocity of the particle is increasing on the interval (0,3)? It also says t is greater than or equal to 0 and gave me its position as x(t)=(t+1)(t-3)^3????????

2. The velocity is the derivative of the position function. Differentiate $x(t)=(t+1)(t-3)^3$ with respect to t.

3. Originally Posted by jinxy33
How do I find the values for t if the velocity of the particle is increasing on the interval (0,3)? It also says t is greater than or equal to 0 and gave me its position as x(t)=(t+1)(t-3)^3????????
Since x(t) is the position of the particle at time t, then the derivative x′(t) is the velocity of the particle at time t. Furthermore, $\frac{d}{dt}x^{'}(t)$ is then equal to the acceleration of the particle. In this case, it appears the question is asking you to find where the accelaration of this particle is positive. So solve for the second deriv of x(t) and study its sign as t varies between 0 and 3.