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?

  • 5
    $\begingroup$ 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. $\endgroup$ – 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.

Velocity is positive, acceleration is negative

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.

The same would apply to the converse as well -- a positive acceleration and a negative velocity would mean a graph which is decreasing ever more slowly, because the velocity is still negative but is increasing. So the "graduating effects" are caused by opposite signs.

Conversely, when velocity and acceleration have the same sign, the graph either "shoots up" or "shoots down", which is what you are seeing here: same signs

Either the velocity is increasing increasingly, or it is decreasing increasingly, depending on whether the signs are both positive or negative (this is a graph of positive signs).

The key point is that speed doesn't have regard for direction -- so whether or not the particle is going left or right is not of your concern. Hence the $|v|


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.