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A particle moves with position function$ s = t^4 - 4t^3 - 20t^2 + t \space \space \space \space t \ge 0 $
(a) At what time does the particle have a velocity of 20 m/s?(b) At what time is the acceleration 0? What is the significance of this value of $ t $?
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01:24
Amrita Bhasin
Calculus 1 / AB
Chapter 3
Differentiation Rules
Section 7
Rates of Change in the Natural and Social Sciences
Derivatives
Differentiation
Missouri State University
University of Michigan - Ann Arbor
University of Nottingham
Idaho State University
Lectures
04:40
In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.
44:57
In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.
0:00
A particle moves with posi…
01:37
particle moves with positi…
02:09
s(t) =t4 4t3 20t2 + 20t (t…
03:21
particle HOTCS with positi…
02:44
A particle moves along a l…
07:05
$s(t)$ is a position funct…
02:26
The position function of a…
01:43
The location of a moving p…
02:10
The acceleration of a part…
01:47
here we have the position function for a particle and we want to know the time when the particle has a velocity of 20 meters per second. So the first thing we need to do is find the velocity function by taking the derivative of the position. So we get fourty cubed minus 12 T squared, minus 40 T plus 20. Okay, we want to set that equal to 20 and solved for the time. 40 cubed minus 12. T squared minus 40 T plus 20 equals 20. We can subtract 20 from both sides and we have fourty cubed minus 12 t squared minus 40 t equals zero. And then we can factor out a fourty from all the terms. It's with fourty times t squared minus three. T minus 10 equals zero, and we can factor the quadratic expression into T minus five times T plus two. Now we have three factors to set equal to 0 40 equals zero T minus five equals zero and t plus two equals zero. We saw each of those and we get t equals zero. We get t equals five and we get t equals negative too. And for this problem we're only using T is greater than or equal to zero. So eliminate t is less than r t equals negative too. Okay, so these are the times when the particle has a velocity of zero zero seconds and five seconds. And for part B, we want to find the time when the acceleration is zero. So we need to take the derivative of the velocity to give us the acceleration. And that would be 12 t squared minus 24 t minus 40. And let's set that equal to zero and solve for T. So 12 t squared minus 24 t minus 40 equals zero. The entire equation is divisible by four, and that would give us three t squared minus 60. Minus 10 equals zero. And unfortunately, that is not factory herbal. So we're going to use the quadratic formula. We get t equals 6/6, plus or minus the square root of 36 minus four times three times negative, 10 over six. Can you simplify that? And you get it down to the point where you have one plus or minus square root 39/3, which you could also write as three plus or minus square 39 all over three. Now, if you evaluate the one with the minus, you're going to find that it turns out to be a negative time. So eliminate the negative time and just keep. T equals three plus square 39/3. That's the time when the acceleration is zero. Now that's very likely to be the time when the particle is changing from speeding up to slowing down or slowing down to speeding up.
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