WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I am Dan Fullerton and in this lesson we are going to start our study of kinematics, looking at the descriptions of motion.
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Our objectives include understanding the general relationships among position, velocity, and acceleration for the motion of a particle.
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Using kinematic equations to solve problems of motion that constant acceleration and that will be carried over into the second half of our lesson.
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Writing an appropriate differential equations and solving it for velocity in cases when acceleration is a specified function of velocity and time.
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Position vs. Displacement.
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An objects position is its location at some given point in time.
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The vector from the origin of the coordinate system to the objects position is known as the position vector which is r.
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Sometimes it is written as rs we are going to use interchangeably in here.
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If an object moves, its position changes.
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This change in position is called displacement δ r or δ s.
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Position and displacement are both vectors.
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They have a direction as well as a magnitude.
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In one dimension position is given by the x coordinate and the displacement oftentimes written as δ x.
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A couple examples about the differences between these.
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A deer walks 1300m E to a creek for a drink.
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The deer then walks 500m W of the berry patch for dinner.
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Before running 300m W it is startled by allowed angry fierce nasty evil raccoon.
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What distance did the deer travel?
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The distance the deer traveled 1300 + 500 + 300 = 2100m but what is the deer's displacement?
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That is where how far it is from that starting point.
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It went 1300m E and 500m W now it is only 800m E and then it went 300m W.
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It is 500m E from where it started.
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Δ x is its displacement is 500m E and because it is a displacement, it is a vector it needs a direction as well.
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Average speed is the distance traveled divided by the time it took to travel that distance.
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¯V oftentimes depicts average speed which is distance travel ÷ time.
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Average speed is a scalar and is measured in m/s.
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Remember that speed S is a scalar S.
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If we look at velocity on the other hand, the velocity is the rate at which position changes
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and position as a vector so the rate at which it changes is also a vector velocity.
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Now average velocity is the displacement during a time interval divided by the time interval, not the distance.
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The displacement is average velocity.
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Average velocity is a vector, it has a direction but it also has units of m/s.
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They are awfully easily to confuse taking the same symbol V.
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Velocity V is a vector, speed S is a scalar S.
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Let us take a look.
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The deer walks 1300m E to the creek, 500m W to the berry patch before running 300m W when it is startled by that loud evil angry nasty raccoon.
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The entire trip took 600s or 10 min.
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What is the deer’s average speed?
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V average is distance over time it traveled 2100m was its distance / 600s.
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The average speed was 3.5 m/s.
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The deer’s average velocity however average velocity is δx /t which was 500m E.
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Its displacement divided by the time 600s or 0.83 m/s E.
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Notice how subtle these are in their differences.
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Average speed you are worried about distance.
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Average velocity you need displacement.
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Let us do an example with our dear friend Chuck the hungry squirrel.
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Pork chop travels 4m E and then 3m N in search of an acorn.
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The entire trip takes him 20s.
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Find how far Chuck travelled.
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That is easy.
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The distance travelled is 4m + 3m which is 7m.
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Chuck's displacement is however is a little trickier.
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Chuck traveled 4m E and 3m N.
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Displacement for a straight line distance from where you start to where you finish so that is a 345 triangle that must be 5m.
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Chuck’s displacement is 5m NE and if we wanted to we can find the angle to be even more specific
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that would be the inverse tan for calling that angle θ of our opposites over the adjacent 3m /4 m which is about 36.9°.
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Chuck's average speed or average speed is going to be distance ÷ time or 7m /20 s which is 0.35 m/s.
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Chuck’s average velocity is a little bit different δx/t how far his displacement divided by time?
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It was 5m NE/ 20s which can be 0.25 m/s and it is a vector.
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It needs a direction NE.
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Subtle difference is you really have to know what you are talking about and read carefully
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when you get in the distance displacements being velocity considerations.
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Let us take a look at acceleration that is the rate at which velocity changes.
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Acceleration is change in velocity overtime.
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It is a vector and it has a direction and the units of acceleration are m/s or m/s².
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That can be confusing to a lot of folks the first time you see it.
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If velocity changes you may go from a velocity of 10 m/s to 20 m/s.
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If that time it takes you to change your velocity by 10 m/s is 1s you went from 20 m/s to 10 m/s to 20 m/s in 1s.
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Your acceleration was 10 m/s every second or 10 m/s².
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As we are talking about the relationship to velocity and acceleration is the derivative of velocity with respect to time or would be written as V prime.
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Acceleration problem.
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Monty the monkey accelerates from rest to velocity of 9 m/s and a time span of 3s.
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Find Monty’s acceleration.
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Acceleration is change in velocity ÷ time = final - initial velocity over time which is 9 m/s - 0 m/s all in 3s or 3 m/s².
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Let us take a look at the position vector.
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If we have a half in space and we have a particle that is moving along that path we could define its position of various points in time.
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Its first position at some time t1 we can define its position by a vector from the origin to that point.
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Let us call that the position vector at time t1.
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A little while later, it is over here at time t2 so we can define the position vector at t2 as such.
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There is its position vector at time t2.
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What happened between t1 and t2?
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That was when we had a change in position δr which would be that position at t2 - the position at t1.
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If we could define S going from that point to that point there is δr.
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While we are doing this we can look at the position function.
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The function of time and realize that our x value changes as a function of time so we have x(t)
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and the I ̂ direction and the unit vector direction along the x axis + y value is a function of time in the y direction, the unit vector in a y direction J ̂ .
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If you prefer you could write this as x is a function of time for the x coordinate, y is a function of time for the y coordinate.
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Of course you can extend at the three dimensions as well.
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Then our average velocity vector is just going to be change in r over some time interval.
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If we take that and go further in the average velocity, we have a particle traveling along the path find the average velocity the between 1 and 6 seconds?
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The average velocity of the change in position divided by time before looking in the x that is going to be x - x is 0/t or we are just traveling in one dimension.
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It is a time vs. distance traveled graph which is going to be our final value to about 5m - 2.5m/ 6s it is about 5 or 6s – 1s or 2.5/5 which is going to be about 0.5 m/s.
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Interestingly though we could also look at the slope here.
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If we go when we try and take the slope for those two points, we will pick a couple points on our graph.
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It looks like an easy one to pick will be 0, 2 and we will also go over here to 6s and say that we are at 5.
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Our slope is rise over run is going to be 5m -2m/6s -0 s or 3m/6s is 0.5 m/s.
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The slope of the position time graph gave us the same thing as velocity it does give you the velocity.
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Looking at instantaneous velocity.
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Average velocity observed over an infinitely small time interval.
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As you make that time interval smaller and smaller until it becomes infinitesimally small you get the instantaneous velocity at that exact point in time.
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Instantaneous velocity is the derivative of position with respect to time.
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Velocity as a function of time is the derivative of position with respect to time or you could write that S prime.
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We wanted to know what the absolute instantaneous velocity was here 3s,
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we would find tangent to the curve the slope right at that point in time.
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That is 2 m/s is our slope that is the instantaneous velocity at that point in time.
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Now we also have the xt graph, the area under velocity time graph is the displacement during that time interval.
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If we make this instead we go to a velocity time graph.
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This shows our velocity is a function of time.
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The area under it, if you integrate that integral of velocity or just find the area that your change in position for that time interval.
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Acceleration is the rate which the velocity changes.
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Acceleration is the limit is Δ t goes to 0 and Δ v/Δ t.
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Usually make that time interval shorter and shorter or the derivative of velocity with respect to time.
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Since velocity is a derivative of position that is the 2nd derivative of the x respect the time.
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And note that we write that as d² x/dt² really squaring anything this is talking about the 2nd derivative.
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The derivative and then to take the derivative of the first derivative.
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The average acceleration is Δ v/t so the velocity time graph we could take the slope here and I get something that looks kind like this I think take the slope at that point.
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If we take the slope of that line, the slope of the velocity time graph at a specific point that would be about -.18m/s².
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Which means that the acceleration at time t= 4s right where that point lines up would be -.18 m /s².
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The slope gives you the acceleration at that point in time.
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We can also look at some graph transformations.
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If you have a position time graph and you take the slope you can get a velocity time graph.
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Take the slope of the velocity time graph you can get an acceleration time graph.
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Or going the other way start with acceleration time graph, if you take the area under the graph you get the change in velocity.
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If you have a velocity time graph and you take the area of the integral you get the change in position.
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You can go from one graph to another based on what you are trying to find using slopes and areas, derivatives, and integrals.
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Alright the velocity, acceleration in two dimensions.
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Our velocity vector is the limit as Δ t approaches 0 with a time interval gets infinitesimally small.
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Δ r/Δ t or we wrote that as the dr dt.
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The derivative of r with respect to time and if r is in multiple dimensions and that would be the x component derivative of the x component in the x direction
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+ derivative of the y component in the y dimension for however many the dimensions you might have.
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Or in bracket notation the dx dt for the x, dy dt for the y.
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Acceleration then is the derivative of velocity with respect to time which would be the 2nd derivative of x with respect to t in the x direction
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+ the 2nd derivative of y with respect to t in the y direction or in bracket notation again t ⁺2x/dt², 2nd derivative of y with respect to t.
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You can keep expanding upon that for however many dimensions you need.
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Typically we are going to be working with 2 and 3 dimensions in this course.
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Let us talk a little bit more about derivatives.
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If we have some function x as a constant × t to some exponent N, the derivative of x with respect to is that power N × our constant × t to the N – 1.
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The basic polynomial derivative formula.
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We are going to be using that a bit so let us practice for second.
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The position of the particles as a function of time is 2 -40 + 2 t² -3 t³.
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Find a velocity and acceleration of the particles as a function of time.
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Velocity is a function of time is the derivative of x with respect to the time or you could write it as x prime.
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You might even see that written as x with dot over it.
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That is going to be, I am going to start at this side just because I like the bigger exponent first that will be -9t²+ 40 -4.
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To find our acceleration as a function of time that is the derivative of velocity with respect time.
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Or the 2nd derivative of x with respect to time or we could write this as V prime or V with the dot,
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or x double prime or x with two dots they all mean the same thing.
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Eventually we take the derivative of our velocity and I would come up with -18t + 4.
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Alright more examples, an object moving in a straight line has a velocity V in m/s that varies with time t.
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According to this function 3 +2t², find the acceleration of the object in 1s.
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Acceleration is just the derivative of the velocity with respect to time which is going to be derivative to the velocity is going to be 4t.
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Since we know in this problem that T = 1s acceleration is just going to be equal to 4 × 1 or 4m/s².
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For part B, determine the displacement of the object between t = 0 and t = 5s.
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We want to know the displacement Δ x that is going to be the integral from T = 0 to 5s.
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We can start using definite integrals here because we are given some limits on the time.
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Our velocity with respect to time which will be the integral from T = 0 to 5s of 3 +2t² Tt which will be 2t³ /3 + 3t all evaluated from 0 to 5s.
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Remember what this means, that means we are going to plug the 5 in for the t first so we would get 2 × 5³ /3³ /3 + 3 × 5 – 0³ + 3 × 0 = 0.
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What I come up with here was Δ x is going to be 5 × 5 = 25 × 5 = 125 × 2= 250 ÷ 3 + 15= 98m.
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There we go the displacement of the object between 0 and 5s.
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Taking a look at another example.
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The velocity of time curve for the tortoise and hare traveling the straight line is shown below.
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I will color the tortoise here in orange and our hare in blue.
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What happens at time T= 30s?
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Interpreting these can get a little tricky so let us take our time and go right through it.
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At T=30s it looks like the tortoise and the hare have the exact same value for speed.
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A tortoise and hare have the same speed.
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How do you know the two have travelled the same distance at time T = 60s?
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Let us see.
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At T=60s if we look here we are given the velocity time graph.
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If we want to know distance travelled that is the area under the graph.
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For the tortoise, that would be the area of this rectangle which is 60 × 4 =240m.
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For the hare, it is the area of this triangle which you can probably see visually is the same or use ½ base × height you can find that the area for the hare is 240m.
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At T =60s, the area under each curve is the same so Δ x must be the same.
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What is the acceleration of the hare at T= 40s?
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What I would do is I would go over here to T=40s and take a look and say the slope of the line there should give you the acceleration.
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Acceleration is slope or change in velocity over change in time that looks like we are going from the slope of that line is -8 m/s / 60s is -4/30 – 2/15m/s².
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That would be the hare’s acceleration.
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One last one, which of the following pairs of graphs best shows the distance travelled vs. time in speed
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and speed vs. time for our car accelerating down the hill from rest?
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Let us take a look at the first one.
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If distance and time, it looks like it is moving the same amount every time and the speed is constant.
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If it is accelerating that does not make any sense, it cannot be A.
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It looks like the further it goes, its distance travelled for each of the time is getting bigger and bigger.
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The slope of this with the different points gives you your speed graph getting bigger at a constant rate.
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B looks like it is accelerating your speed is constantly increasing at a constant rate.
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This does not make sense because the slope of this does not match your speed curve.
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Same here the distance is increasing linearly while speed is going up quadratically.
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The answer must be B.
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Hopefully that gets you a start on describing motion.
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We will go a little bit further with this in our next lesson describing motion part 2.
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Thank you so much for watching www.educator.com and make it a great day.