Consider a particle moving along the x -axis where x(t) is...

Consider a particle moving along the $x$ -axis where $x(t)$ is the position of the particle at time $t, x^{\prime}(t)$ is its velocity, and $x^{\prime \prime}(t)$ is its acceleration.
Repeat Exercise 61 for the position function
$x(t)=(t-1)(t-3)^{2}, \quad 0 \leq t \leq 5$

Key Concepts

Position Function

A position function describes the location of a particle along a line as a function of time. It serves as the foundation for analyzing motion, providing a starting point from which derivatives can be used to extract further information about the particle's movement.

Velocity

Velocity is the first derivative of the position function with respect to time. It indicates both the speed and the direction of the particle, and is used to determine intervals of increasing or decreasing position and to locate critical points like turning points in the motion.

Acceleration

Acceleration is the second derivative of the position function, representing the rate of change of velocity with time. It helps in understanding how the particle's speed is changing—whether it's speeding up or slowing down—and plays a key role in identifying inflection points in the motion.

Critical Points

Critical points occur where the first derivative (velocity) is zero or undefined. These points are important because they can indicate local maxima or minima in the motion, such as points where the particle changes direction, and are essential for analyzing the overall behavior of the particle.

Transcript

00:05 So for this question, we will be answering a few questions about a particle defined by the following position function.

00:16 We can go ahead and actually foil this out just to make it a little bit easier to perform derivatives on this.

00:24 So, t minus 3 squared first, t squared minus two times the last number plus the last number squared, then go ahead and multiply this out.

00:57 And if we simplify this and combine like terms, we get t -tubed minus 7t squared plus 15t minus 9 as our position function.

01:09 So the first question we have here is to find the velocity acceleration functions of the particle.

01:18 We know that the velocity of a particle is equal to the derivative.

01:23 The first derivative of its position, and we know that acceleration is equal to the first derivative velocity.

01:32 Using that, we can say that velocity is equal to the first derivative position, which will do with the power rule.

01:40 So 3, bring down the 3, t squared, minus 7 times 2, 14, t plus 15.

01:48 We can actually factor this out a little bit, which isn't necessary, but it could be helpful to 3 t minus 5 times t minus 3 now we can go ahead and do the acceleration by doing the first derivative of velocity so a at t and again we're just using the power roll here is 3 times 2 6 t minus 14 and again the the rule we just followed right there is that d of d t of t to the end is equal to a t to the a minus one so our acceleration function is 60 the 14th can be somewhat factored by taking the greatest factor of two two times three t minus seven okay so the next question that they have for us is to find the open t intervals in which the particle is moving to the right and we can say that particle be particle moving right means velocity is positive and we know that because velocity determines the speed of a particle and the cyanide velocity determines the direction in which that speed is aimed the speed is aimed in the positive direction and we consider positive to be to the right then the particle will be moving to the right whenever the velocity is positive so using our velocity function and we're going to use the factored version, just to make things a little easier.

03:41 We can set it equal to zero to determine the velocity is zero, and then use a number line to determine when the function is actually positive.

03:50 So v.

03:51 T equals zero, t equals five -thirds, and t equals three.

04:01 And put those on a number line...

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