# Speeding Up and Slowing Down of Particle

If the position function of a particle is $s(t)=t^3-12t^2+36t$ and t belongs to [0, 10], where t is time in seconds and s is position in meters... At what times is the particle speeding up/slowing down?

I know velocity is the first derivative: $V(t)=3t^2-24t+36$ and acceleration is second derivative $A(t)=6t-24$.

Does speeding up mean a positive acceleration and slowing down mean negative acceleration?

Why is it to find if the particle is speeding up/slowing down, you need to multiply the velocity function with the acceleration function and find the intervals where it is positive/negative?

• The speed is the absolute value of the velocity. Speeding up means the speed is increasing, which means the acceleration and the velocity have the same sign. Slowing down means the speed is decreasing which means the acceleration and velocity have opposite signs. – Spencer Oct 28 '15 at 0:13

Velocity is a vector quantity, and indicates both speed (by its slope) and direction (by its sign). Speed is a scalar quantity, and represents, colloquially, how "fast" the particle is moving (distance over time). And because it doesn't matter in which direction the particle is moving, speed is represented by $|v|$. As Spencer commented, when velocity and acceleration are both positive or both negative, the speed is increasing. When they are different signs, then the speed is decreasing.
To see why, look at this portion of the graph of $x^3$ as x approaches 0. The particle's graph is going up for sure (positive velocity). However, the rate by which its increasing is decreasing (negative acceleration) -- hence why its increasing ever more gradually. In other terms, it's slowing down, because negative acceleration indicates a decreasing velocity.