If a rock is thrown upward on the planet Mars with a velocity of 10 m / s, its height in meters

If a rock is thrown upward on the planet Mars with a...

If a rock is thrown upward on the planet Mars with a velocity of $10 \mathrm{m} / \mathrm{s},$ its height in meters $t$ seconds later is given by $y=10 t-1.86 t^{2}$ $$\text{ (a) Find the average velocity over the given time intervals:}$$ $$ \begin{array}{ll}{\text { (i) }[1,2]} & {\text { (ii) }[1,1.5]} & {\text { (iii) } [1, 1.1]} \\ {\text { (iv) }[1,1.01]} & {\text { (v) }[1,1.001]}\end{array} $$ $$\text{ (b) Estimate the instantaneous velocity when $t=1$ }$$

If a rock is thrown upward on the planet Mars with a velocity of $10 \mathrm{m} / \mathrm{s},$ its height in meters $t$ seconds later is given by $y=10 t-1.86 t^{2}$
$$\text{ (a) Find the average velocity over the given time intervals:}$$
$$
\begin{array}{ll}{\text { (i) }[1,2]} & {\text { (ii) }[1,1.5]} & {\text { (iii) } [1, 1.1]} \\ {\text { (iv) }[1,1.01]} & {\text { (v) }[1,1.001]}\end{array}
$$
$$\text{ (b) Estimate the instantaneous velocity when $t=1$ }$$

Key Concepts

Limit Process

The limit process involves the concept of approaching a particular value as the independent variable gets arbitrarily close to a certain point. In calculus, it is used to precisely define instantaneous rates of change (the derivative) by taking the limit of the difference quotient as the interval shrinks to zero.

Average Rate of Change

The average rate of change of a function over an interval is calculated as the difference in the function's values at the endpoints of that interval divided by the length of the interval. This is essentially the slope of the secant line connecting the two points on the function's graph. It gives an overall measure of how the function is changing over that period.

Instantaneous Rate of Change

The instantaneous rate of change represents the slope of the tangent line at a specific point on a function's graph. It is defined as the limit of the average rate of change as the time interval approaches zero. In physical context, such as motion, it corresponds to the instantaneous velocity.

Derivative

A derivative is the result of computing the instantaneous rate of change of a function with respect to one of its variables using the limit process. It is a fundamental tool in calculus for understanding and modeling the behavior of functions, particularly how they change at any given point.

Transcript

00:01 So we're trying to find average velocity over four different time intervals.

00:06 And we know that velocity is equal to the displacement of x over the displacement of t.

00:17 So in order to solve this problem, we're just going to have to find all of the displacements of distance and the displacements of time for those five different time intervals.

00:27 And to do this, i'm just going to make a table.

00:31 With the first one being the second time, t2, the first column being t1.

00:41 The third column is going to be the displacement of them.

00:46 That's just by subtracting t2 from t1.

00:51 And then we will have y2, y1, and then again the displacement of y, which will be done by taking y2 minus y1.

01:10 And it's y in this case because in the context of our point, problem.

01:15 We have a rock going up in the air, which would be up and down, which is y on the graph.

01:24 So in this problem, v is, can be written as v is equal to the displacement of y or the displacement of t.

01:31 And then now that we have the displacement of y and the displacement of t, the last column will just be solving by dividing the two.

01:39 That would be the philosophy.

01:41 So for the first interval, we're given 2 and 1, then 1 .5 in 1, then 1, and 1, 1 .1 and 1 .01 and 1 .01 and 1.

02:00 And finally 1 .001 and 1.

02:06 So the displacement for all of these would just be by subtracting the left column from the right column.

02:11 And those would give us 1 .5 .1 .1 .01 and point 0 .1.

02:24 And now to find y2, we just plug in the t to into the equation given, that is 10 t minus 1 .86 t squared.

02:37 And this would give 12 .56, 10 .815, 8 .75, 8 .75, 8 .2, 8 .25, 8 .2, and 8 .15.

03:13 And then y1 will be the same thing, but with plugging in 1.

03:17 So all of these are going to be the same as they all have the same t1 value, and that is 8 .14.

03:32 So then to find the displacement of y, we'll just subtract all the y2s from all the y1s.

03:37 This will give 4 .42, 2 .675.

03:51 These values won't match up perfectly if you plug them in using purely what i have here...

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