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...

Key Concepts

Physical Interpretation of Velocity

This concept involves understanding the significance of both the magnitude and the sign of velocity. In projectile motion, a positive velocity typically indicates upward movement, while a negative velocity indicates downward movement. This interpretation is crucial for analyzing the object's behavior as it moves through its trajectory.

Quadratic Equations

A quadratic equation is a second-degree polynomial equation that can usually be solved using methods such as factoring, completing the square, or the quadratic formula. In projectile motion problems, quadratic equations are often used to determine the time at which an object reaches a particular position, such as when it returns to the surface.

Projectile Motion

This concept involves the study of objects that are thrown or projected into the air and are subject only to the acceleration of gravity. It incorporates the analysis of the object's position, velocity, and time of flight using kinematic equations and is fundamental in understanding the motion of the rock.

Differentiation

Differentiation is a key concept in calculus that deals with finding the rate at which a function changes at any given point. In the context of motion, differentiating the height function with respect to time gives the velocity function, which helps analyze the speed and direction of the moving object.

Transcript

00:01 So before we actually do a, i would make note that it would probably be better to do b first and then just use the result from b to plug in a.

00:13 But the way they have it set up, they want us to do this first and then to go from there.

00:17 So i'm just going to go ahead and follow the steps.

00:20 But if you want, you can just kind of skip ahead, see what we do in b, and then just plug one into that value.

00:26 And that should give you the same thing to get over here.

00:28 I mean, it'll save you a lot of time, but just kind of first.

00:30 Sake of how they have the problem set up.

00:32 I'm going to kind of go from a than b.

00:36 So over here, remember that the velocity, if we think about it, is just the derivative of our position.

00:44 So in this case, h prime of t is going to be equal to v of t.

00:51 And we know, and maybe i should write out what the definition of this is first.

00:59 So h prime of t is going to be equal to the limit as h approaches 0 of h of t plus h minus h of t all over h.

01:18 And so in this case, if we want one, then we're just going to replace those t's with ones.

01:25 So since this is the same thing as v of t, we could just go in and write that.

01:29 So it would be v of 1 is equal to the limit as h approaches 0 of capital h of 1 plus little h minus h of 1 all over little h.

01:46 And now we can plug in 1 plus h into here as well as 1.

01:51 So that would be the limit as h approaches 0.

01:55 So h of 1 plus h, so that would be 10 times 1 plus h.

02:02 And then minus 1 .86 1 plus h squared, minus, well if we come over here and plug in 1, so that would be 10 minus 1 .86.

02:14 And then this is all over h.

02:17 Actually, i'll need to scoot this down a bit more, it looks like.

02:25 Then we can go ahead and distribute everything.

02:29 So this would just be 10 plus 10h.

02:34 This here is going to be, well, if we expand that out, it would first be 1 plus, 2h plus h squared and then we would distribute the negative 1 .86 so i'll go ahead and do that and actually before we do that notice first this 10 and that 10 cancel out with each other and then if we distribute this negative 8 .6 this one and this will cancel out because we'd have negative there but yeah so those cancel out so actually let's write that out just so you can see what we have left before i actually start doing stuff.

03:09 So it would be 10h minus 1 .86 times 2h plus h squared all over h.

03:18 And then we can go ahead and simplify this down by combining our like terms.

03:25 So we'd have 10 minus.

03:30 Actually, let me distribute that 1 .86, so 1 .86 times 2.

03:34 So this is going to be minus 3 .72, and then minus 1 .86 squared.

03:45 There should be an h there.

03:46 Then we can combine those, so 10 minus that would be 6 .28.

03:51 So we have the limit as h approach to 0 of 6 .28h minus 1 .86h squared all over h.

04:02 And now notice, we can divide this h into each of those terms.

04:05 To just give the limit as h approaches 0 .0 .6 .28 minus 1 .86h...