WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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In this lesson we are going to continue to talk about describing motion and kinematics.
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Our goals understand the relationship among position, velocity, and acceleration for the motion of the particle.
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Use the kinematic equations to solve problems of motion with constant acceleration
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and write appropriate differential equations and solve for velocity in cases in which acceleration is specified function of velocity and time.
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Let us start talking about constant acceleration.
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For cases where we have a constant acceleration, where velocity is always changing by a constant rate or not changing at all,
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we can derive a set of kinematic equations that will allow us to solve for unknown quantities.
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These pre key equations are the final velocity = the initial velocity + acceleration × time.
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The final position is the initial position + the initial velocity × time + half the acceleration ×
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the square of the time that has elapsed or the square of the final velocity = the square of the initial velocity +
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2 × the acceleration × the change in position.
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When you know any of these 3 kinematic quantities with constant acceleration you can use these formulas to solve for the other 2.
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The trick is always going to be finding and understanding what 3 of these are before you go on to get the rest of them.
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How did we come up with these equations?
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Let us take a minute and see if we can derive them.
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Here we are showing a velocity time curve for something that is steadily increasing its speed.
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It is accelerating at a constant rate.
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To find our kinematic equations I'm going to start with the definition of acceleration a = Δ V / T which is our final velocity - our initial velocity ÷ time.
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Or I can rearrange that to say that the final velocity = initial velocity + acceleration × time.
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There is our first kinematic equation.
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For our second, we are going to take a look at this in a little more detail and pull out my ruler here.
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I'm going to look at this where we have this as a function of time.
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We are going to call this for some final value time T and I'm going to break up our shape here into a triangle and
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a rectangle that is going to come in handy in just a minute.
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As we look at our graph, if this is our initial velocity, this is our final velocity, then this distance here is V – V initial.
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If we want to take a look at the change in position Δ x that is the area under the velocity time curve.
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Δ x is the area just ½ base × height for our triangle + length × width our rectangle down here
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which is going to be ½ , our base is T and our height of our triangle is V - V initial + our height V knot × T.
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But we already know from equation 1 that V =V initial + at.
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Therefore V - V initial must equal at.
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This implies then that Δ x = ½ T × V – V initial, we are just going to replace with at + V knot T which implies then that Δ x must equal V knot t + ½ at².
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There is equation 2.
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For our 3rd equation, we are going to come back up to the formula we have there in green, Δ x = ½ T × V – V knot + V knot t we have right here.
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We are just going to manipulate it a little bit differently.
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We are going to say then that ½ Vt – ½ V knot t + V knot t which implies then that Δ x = ½ Vt + ½ V knot t
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which implies then that Δ x / t = V + V knot / 2 which is another formula not exactly whenever kinematic equations but useful.
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That left hand side is V average and V average is V final + V initial ÷ 2 for something that is accelerating that has a constant acceleration.
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We are going to use that here in a little bit, I believe.
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Let us keep going with derivation now.
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V average = Δ x / t which implies then that Δ x = V average × t which we just said is V + V knot / 2 × t.
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Another tricky little substitution here if V = V initial + at, we can solve that for t to say that t = V - V initial / a.
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We can then substitute in to say that Δ x = we have our V + V knot / 2.
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For our t, we have V – V knot / a which implies then if we multiply through here Δ x is going to be V² - V initial² / 2a
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which implies then that V² - V initial² = 2a Δ x or solving for V² V final² = V initial² + 2a Δ x.
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There is that 3rd kinematic equation.
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It is fairly easy to derive these.
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As we start dealing with a bunch of these kinematic problems, there are couple problem solving steps
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when you can use kinematic equations for constant acceleration problems that will help things out.
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First is to label your analysis for horizontal and vertical motion.
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Choose a direction to call positive and stick with that throughout the problem.
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Typically it helps if you call the direction the object is moving initially as the positive direction.
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Create a motion analysis table.
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Fill in your givens and once you know 3 items in your table you can always solve for the unknowns using your kinematic equations.
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Let us take a look at an example here.
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A race car starting from rest V initial = 0 accelerates uniformly at a rate of 4.9 m / s² that must be your acceleration.
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What is the car's speed?
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We are looking for V final after it has traveled 200m Δ x.
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Let us pick a direction to call positive and we will call to the right positive in this case.
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Horizontal motion problem and we will list our items of interest, V initial, V Δ x, a, and t.
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We will fill in what we know, V initial is 0, we are trying to find V.
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The final Δ x is 200 m, acceleration is 4.9 m / s², and we do not know time.
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What I am going to do now is I'm going to look at what I have and what I'm trying to find
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and see if I can find a kinematic equation that has all the items that I want in there.
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I do not care about t, what I am looking for a formula that has those things in it.
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There is one that would be V² = V initial² + 2a Δ x.
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V² = 0² + 2 × 4.9 m / s² × Δ x 200m or V² = 1960 m² / s².
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I do not want V², I want V.
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V is going to be equal to the square root of that which is + or - 44.3 m / s.
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Common sense tells me of course I'm looking for the positive so I can get rid of the negative there.
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My answer is 44.3 m / s.
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Let us take a look at a vertical problem.
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An astronaut is standing on the platform on the Moon drops a hammer.
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If the hammer fall 6 m vertically in 2.7 s what is its acceleration?
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This is a vertical problem and since the hammer goes down first that is called down our +y direction.
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Then the list of our information of interest V knot, V, Δ y, a, and t.
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We know V initial is 0, we know Δ y is 6m and there time is 2.7 s.
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What we are trying to find? Acceleration.
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Our 4 items that would be nice to have all a nice formula are right there.
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We can do that using the formula Δ y = V initial t + ½ at².
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V initial is 0 that term becomes 0.
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This becomes a little simpler Δ y = 1/2 at² which is ½ .
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Actually we are trying to solve for a so this implies then that a = 2 Δ y / t² which is 2 × 6 m / 2.7 s² or about 1.65 m / s or 1.65 m / s².
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How about a problem that is a little bit more involved, a 2 step problem?
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We have a car traveling on the straight road at 50 m / s and it accelerates uniformly to speed of 21 m / s in 12 s.
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Find the total distance traveled by the car in that 12 s interval.
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We will call to the right our positive direction and list our items of interest V knot = 15 m / s, V final is 21 m / s, Δ x is what we are trying to find.
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a we do not know and time is 12 s, that is what I'm after.
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However, it does not look to me like we have a very straightforward path to getting that as it currently stands.
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What formula has those 4 things in it I do not see it.
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What we could do know is we could remember the average velocity is V initial + V final / 2.
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That is going to be 15 m / s + 21 m / s / 2 or 18 m / s.
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Average velocity is distance traveled ÷ time, therefore the distance traveled is going to be V average × time or 18 m / s × 12 s which is 216 m.
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One way you can go about solving that.
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We do not actually have to do it that way especially if you do not remember that equation.
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We could also go and we can find acceleration first.
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We know enough to find acceleration even though we do not need it and then plug that back into another kinematic equation.
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Let us try that a = change in velocity / time which is final velocity - initial velocity / t or 21 m / s - 15 m / s ÷ 12 s.
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It is just going to be 0.5 m / s², we know a = 0.5 m / s².
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We can choose any kinematic equation we want now because we know the 4 other items as long as it has Δ x in it.
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Let us choose Δ x = V initial t + ½ at² which will be 15 m / s × 12 s + ½ × the acceleration 0.5 m / s² × 12 s² is going to give us 216 m again.
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Or we could have picked even a different kinematic equation, V final² = V initial² + 2a Δ x.
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Solving for Δ x that is going to be V final² - V initial² / 2a which will be (21 m / s² -15 m / s² ) / 2 × 0.5 m / s² or 216 m.
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A lot of different path to a solution.
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Something we are going to see coming up again and again and again in physics.
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There may not be any correct single path, there may be multiple paths to an answer.
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Some may be considerably easier than others or more straightforward but oftentimes
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there more than one solution to a problem that will lead you to a correct answer.
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Taking a look at acceleration.
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How long must a 5 kg kitty cat accelerate at 3 m / s² in order to change its velocity by 9 m / s?
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That looks like a horizontal problem again, V initial, V Δ x, a, and t.
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What do we know?
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We want to have a change in velocity of 9 m / s.
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Δ V = 9 m / s and acceleration is 3 m / s².
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We can do this quite simply if we wanted to recognizing that acceleration is change in velocity /time.
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Therefore time = Δ V / a or 9 m/s / 3m/s² which is 3s.
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Let us take a look at some particle diagrams.
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A spark timer is a little device that shots at regular intervals and what you do is
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you have a paper go through it and it makes dots on the paper.
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Those dots correspond to that time interval so you can examine the dots to help you see what is going on with the motion of an object.
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A spark timer is used record the position of a lab card accelerating uniformly from rest.
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We probably tied this piece of paper onto a lab card and that card go through it and had the card move
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as it went through the timer it makes a dot and set time intervals that gives you a feel for what is going on.
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Each 0.1s the timer marks a dot on a recording tape to indicate the position of the cart.
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If that is a 0s and that means this must been 0.1 s, that dot must have been at 0.2 s, 0.3 s, 0.4 s, you get the idea.
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Find its placement of the card at time t = 0.3 s.
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I will do that and just go to 0.3 s to recognize from my ruler that must be 9 cm, Δ x = 9 cm.
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Find the average speed of a cart from 0 to 0.3 s.
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Average speed is displacement ÷ time so that would be 0.09 m or 9 cm ÷ 2.3 s to get there which is 0.3 m / s.
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Let us go a little further with this problem.
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Find the acceleration of the cart.
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To find the acceleration of the cart we could use Δ x = V knot t + ½ at² realizing that the cart began at rest.
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It says it is accelerating from rest V knot 0.
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Therefore we can rearrange this for a to say that a is going to be 2 Δ x / t² or 2 × it went 0.09 m here at 0.3 s ÷ 0.3 s²
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going to give us an acceleration of about 2 m/s².
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On the blank diagram draw at least 4 dots indicating a cart moving at constant velocity.
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I will probably have something like that constant velocity they should be evenly spaced.
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When you do these problems there are times when you can come into quadratic situations.
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Arnie, the aardvark accelerates at the constant 2 m/s² from an initial velocity of 1 m/s.
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How long does it take Arnie to cross a distance of 50 m?
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Horizontal problem will start with our known information V initial as 1 m/s.
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We do not know V final, displacement is 50m, acceleration is 2 m/s².
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We do not know the time and we are trying to find that.
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As I go to solve this I look for my items of interest, we have got those 4.
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We are worried about that one time.
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We have got a formula that has those in there.
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Δ x = V initial t + ½ at² which implies that Δ x 50 = V initial is 1.
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T + ½ a and ½ × 2 is going to be 1.
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T² which implies then that t² + t - 50 = 0.
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It is a quadratic equation and we can use the quadratic formula.
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Let us do that for practice t = -b + or - the square root of b² – 4ac / 2a which is going to be -b is going to be -1 + or –
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the square root of b² 1² - 4 × a which is 1 × c which is - 50 / 2a which is 1.
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That is going to be equal to -1 + 14.18 / 2 or -1 -14.18 / 2 which implies that t = 6.59 s or -7.59 s.
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Here we got to use a little bit of common sense and know which one of those does not really fit here?
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Of course that is the negative time, the correct answer must be 6.59 s.
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That is a lot of work.
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Typically when you see a quadratic formula pop up in a kinematic equation like this you can get around it by being a little bit clever.
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An alternate solution, imagine instead we want to solve for the final velocity first.
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We could do something like final velocity² = the initial velocity² + 2a Δ x which is going to be equal to 1² + 2 × acceleration 2 × displacement 50,
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which implies that V² is going to be equal to 201 m / s² or V final is about 14.18 m / s.
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Maybe a little bit familiar.
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Once we have that we could go to another kinematic equation V = V initial + at
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therefore t = V - V initial / a which is 14.18 m / s - 1 m/s ÷ 2 m/s².
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All looks familiar perhaps which gives us 6.59 s same answer but we did not have to go to use the quadratic formula.
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And really is just another way of getting to the same thing.
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If you do see you got a quadratic and you are not a big fan of using a quadratic formula solve for something else first
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and then you can go use a different kinematic equation.
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Free fall when the only force acting on an object is the force of gravity the objects weight we refer to the motion of that object as freefall.
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That we are going to ignore the force of friction or air resistance.
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If it drops something it is in free fall, that includes objects that have a non 0 initial velocity.
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If you throw something down as long as the only force acting on it is the force of gravity we will call that freefall.
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If we drop a ball and a sheet of paper of course it is obvious there not going to fall at the same rate.
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If you can remove all the air from the room however you would find that they do follow at the same rate.
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We are going to analyze the motion of objects by neglecting air resistance that type of friction at least for the time being.
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Eventually we are going to pull air resistance back into the mix here in this course.
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Near the surface of earth objects accelerate down at the rate of 9.8 m/s /s we call
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this acceleration the acceleration due to gravity and gets a special symbol g.
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For the purposes of the AP exam we want to make the math a little quicker you can round 9.8 up to 10.
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More accurately g is the gravitational field strength.
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As we move further away from earth of course the acceleration due to gravity decreases.
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Let us take a look at objects falling from rest.
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That object falls from rest that means its initial velocity is 0 since the objects initial motion is down
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it is typically useful to call down the positive y direction and acceleration is +g.
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If we called down the y direction and the object accelerates down that would be a = 9.8 m/s².
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G is always 9.8 m/s² whether the acceleration is +g or -g depends on your reference.
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Let us take a look at falling objects.
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How far will a brick starting from rest fall freely in 3s neglect air resistance.
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Since it is going down first we will call down our y direction.
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We will set up our table of information V initial = 0 from rest.
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V final we do not know, Δ y we do not know.
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Acceleration we do know it is 9.8 m/s² and let us round that to 10 m/s² and that is positive because we call down positive and time is 3s.
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Using our kinematic equations to find how far Δ y = V initial t + ½ ay t²
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and since V initial is 0 that term becomes 0, Δ y = 1/2 a 10 × t² 3².
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9 × 10, 90 × ½ 45m
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Alright how about objects launched upward?
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In this case you must examine the motion of the object on the way up and the way down.
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Since the objects initial motion is up typically it is useful to call that the positive direction.
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If we call up the + y direction the acceleration and the problem would be -g or -10 m/s²
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because it is pointing in the opposite direction of what we called positive.
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It is nice when things coming up and back down is it the highest point when it gets to its highest point,
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that peak for a split second its vertical motion stop since velocity there is 0 before it starts accelerating again on the way back down.
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There is some symmetry of motion there.
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We throw something up and we are neglecting air resistance it takes the exact same amount of time
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to go up as it does become down to that same level.
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The initial velocity you threw it up with is the same as the initial velocity it comes down with.
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Theoretically if you neglect air resistance and you shoot a bullet straight up, the velocity of the bullets as it comes out of the gun
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is going to be the exact same as the velocity when it is coming down just in the opposite direction.
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In reality it does not work that way because of air resistance but I think you get the idea there.
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A ball thrown upward, a ball thrown vertically upward reaches a maximum height of 30m above the surface of the Earth,
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find the speed of the ball and its maximum height.
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Let us call up the positive y direction and we are going to have to recognize as the object goes up and comes back down at its highest point the vertical velocity is 0.
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Therefore the speed of the ball at its maximum height 0 m/s.
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A basketball player jumps straight up to grab a rebound, if she is in the air for 0.8 s which is pretty good jump how high does she jumped?
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Let us call up our +y direction and realize that she comes up and back down and all of that happens in 0.8 s
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which means she hits her highest point at 0.4 s half of that.
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If we wanted to we could analyze her motion on the way up or the way down that breaking this in half is going to make it a lot easier
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because if you want to know how high she jumps we want to know what is happening halfway through that motion.
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Let us look at the way up first.
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Up at y direction we know now V initial, we do not know we know final velocity is going to be 0, Δ y is what we are trying to find.
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A must be -10 because we call the positive in the time to go up is 0.4 s.
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Let us see do we have an equation that will give us Δ y write off?
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I do not see anything easy.
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Let us solve for initial velocity first.
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If a V = V knot + at and we could say that V initial = V - at which is 0 - a -10 × 0.4 s is 4 m/s.
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Our initial velocity must have been 4 m/s.
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We can use an equation with Δ y we want to find how high she goes.
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Her highest point Δ y is V initial t + ½ at² which is going to be V initial is 4 m/s ×
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our time 0.4 s + ½ × acceleration -10 × time squared 0.4²
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Complies that Δ y is about 0.8 m that is a pretty serious vertical leap.
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That was one way to do it.
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We could have also looked at what was going on by analyzing our motion on the way down instead of the path up.
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If we want to look at the way down what we know she starts at initial velocity 0 so it is called down the positive y.
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V initial before looking at just this window where she is coming down the initial is 0.
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We do not know V final, we do not know Δ y.
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Acceleration now is 10 m/s² because she is accelerating down and we called down y direction and t is 0.4 s.
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Δ y = V initial t + ½ at² which is ½ × 10 m/s² × 0.4² or 0.8 m.
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Noticed that analyzing on the way down we only have these one formula so it was a little bit faster.
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Either one will work but another case where there are multiple ways to solve the problem
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but one is typically a little more streamlined than the other.
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How about a ball thrown downward?
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We threw a ball straight down there with the speed of 0.5 m/s from a height of 4m.
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What is the speed of the ball 0.7 s after it is released?
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All right it is called down the y direction because that is the direction the ball moves initially.
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V initial is .5 m/s we do not know V and we want to figure that out.
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Δ y is a little tricky I will put 4 m in there but it does not say the ball travels 4m it says
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it was thrown from a height of 4m and does not say the hits the ground.
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Therefore we do not know how far it is gone when it is 0.7 s.
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From now we do not know the Δ y, A is 10 m / s² and our time is 0.7 s.
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Now V = V initial + at which is 0.5 m/s + a 10 m/s² × 0.7 s which is going to be 7.5 m/s.
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There is our final velocity.
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Alright a quarter kilogram baseball is thrown upward with the speed of 30 m/s.
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Neglecting friction the maximum height reached by the baseball is what?
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Let us take a look and we are going to call up the y direction and as we do that the maximum height all we know V initial is 30 m/s.
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Its highest point, its top, just by looking at just the way up we can call its final velocity 0 m/s.
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Δ y is what we are trying to find, acceleration is -10 m/s² because we called up our positive direction and time we do not know either.
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I'm looking for something that has those quantities in it.
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V² = V initial² + 2a Δ y which implies that Δ y = V² - V initial² / 2a.
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Now we can substitute in our values Δ y = 0² - 30² / 2 × -10 – 900 /-20 is going to give us 45m.
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Another type of popular problem is called catch up problems or something like this.
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Rush, the crime fighting super hero, can run at the maximum speed of 30 m/s while Evil Eddie the criminal mastermind can run 5 m/s.
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If evil Eddie is 500m ahead of Rush, how much time does Evil Eddie have to devise an escape plan?
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Let us see here, we can start by looking at their positions.
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The position of Rush is going to be given by the speed × time.
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Displacement is velocity × time so 30 m/s × time.
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The position of Evil Eddie is he starts 500m ahead of Rush but adds at the rate of 5 m/s 500 + 5t.
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How much time does Evil Eddie have?
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He has until Rush catches him, when their positions are the same.
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If we have let xr Rush’s position = Evil Eddie’s position and solve for time that should tell us how long Evil Eddie has to figure out how to get out of the situation.
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That implies then that 30t = 500 + 5t or 25t = 500 therefore t = 500 / 25 or 20s.
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Evil Eddie better do a lot of thinking in that 20s.
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How far must Rush run to capture Evil Eddie?
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The distance Rush covers is going to be his position, since he started at 0 that is going to be 30t which is 30 × 20 s or 600m.
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Rush is going to run 600 m in the same time Evil Eddie runs is 100m because he started at 500.
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Alright, one more sample here.
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3 A model rocket of varying mass are launched vertically upward from the ground with varying initial velocities,
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from highest to lowest rank the maximum height reached by each rocket neglecting air resistance.
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It looks like they have different masses, from lightest to heaviest.
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This one has the highest initial velocity, the lowest, and medium.
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The key here is there are no mass dependents.
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The only thing that is going to matter is that initial velocity.
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The one with the highest initial velocity with the biggest initial velocity is going to go the highest.
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That will be one first which is going to then go to the rocket 3 the next highest
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and rocket 2 will have the lowest maximum height because there is no mass dependents.
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All right thank you for watching www.educator.com and make it a great day everyone.