Presentation on theme: "Horizontal and Vertical Equations in one dimension"— Presentation transcript:
1Horizontal and Vertical Equations in one dimension KinematicsHorizontal and Vertical Equationsin one dimension
2So far… Displacement → Δx = xf – xi Velocity → v = Δx BASIC EQUATIONS LISTDisplacement → Δx = xf – xiVelocity → v =Acceleration → a =ΔxΔtΔvΔt
3If we know 3 of these, we can figure out the other 2. We now know everything we need to know to predict the path of an object in horizontal motionx t vf vi aIf we know 3 of these, we can figure out the other 2.
4From before we know:This equation can be rearranged to look like:This equation is the 1st kinematics equation.
5If we take the integral of this equation, we get : From this equation we can determine the 2nd equation.If we take the integral of this equation, we get :Most of the time, xi will be 0.This equation is the 2nd kinematics equation.
6For the 3rd kinematics equation: vf = vi + a t t = (vf – vi) / a x = vi t + ½ a t 2 x = vi [(vf – vi) / a] + a [(vf – vi) / a] vf2 – vi2 = 2 a x vf2 = vi2 + 2axMost of the time, xi will be 0.
7Kinematics Formula Summary For 1-D motion with constant acceleration:vf = vi + a tx = vi t + ½ a t 2vf2 = vi2 + 2 a x
8To Remember… If there is acceleration, we need more than just x = vt Because the object has either increased speed, therefore going furtherOr the object has decreased speed, therefore not covering as much ground.
9When do you use each equation? Identify what you want to know.Identify the information given.Eliminate equations based on this information.Example: A catcher catches a 90 mph fast ball. His glove compresses 4.5 cm. How long does it take to come to a complete stop?Looking for timeGiven90mph4.5cmWe know that the ball is stopping
10Example continued vf = vi + at x = vit + ½at2 vf2 = vi2 + 2ax Looking for time = tGiven90mph = vi = 40.2m/s4.5cm = x = 0.045mstopping = vf = 0vf = vi + atx = vit + ½at2vf2 = vi2 + 2axWhich equation do we use? Can we solve with just one?Need to find acceleration first.Use #3.a = m/s2Then find time.Use #1.s or msAnswer
11You try:A dune buggy accelerates uniformly at 1.5 m/s2 from rest to 22 m/s. Find the total distance traveled and the total time.A car is moving at 30m/s when the brakes are applied. It stops 2.5 s later. Find the car’s acceleration and how far it traveled in that time.x =161.3mt = 14.7 sa = -12m/s2x = 37.5m
12Galileo GalileiIn addition to telescopes and his other pursuitsGalileo formulated the laws that govern the motion of objects in free fall
13Free FallAll objects moving under the influence of gravity only are said to be in free fallFree fall does not depend on the object’s original motionAll objects falling near the earth’s surface fall with a constant accelerationGravity accelerates the object toward the earth the entire time it rises, and the entire time it falls.This acceleration is called “acceleration due to gravity,” and is indicated by “g”
14Acceleration due to Gravity g = 9.8 m/s²When estimating, use g » 10 m/s2g is always directed downwardToward the center of the earthIgnoring air resistance and assuming g doesn’t vary with altitude over short vertical distances, free fall is constantly accelerated motion
15Vertical Kinematics Equations Replace a with gand x with yvf = vi + gty = vit + ½gt2vf2 = vi2 + 2gy
16Free Fall of a dropped object Initial velocity is zeroUse the same general kinematics equationsGenerally use y instead of x since verticalAcceleration is g = 9.8m/s2Because the object is speeding up, we will say that g is positiveThis does not agree with your book.vi = 0a = g
17Free Fall of an object thrown downward a = g = 9.80 m/s2Initial velocity 0g is positive because the object is speeding up.We can choose that downward direction is positive when dealing with falling objects soInitial velocity will be positiveThe only thing acting on the object after launch is gravity… FREE FALLvi 0a = g
18What’s your reaction time? Practice Problem: You drop a ball from rest off a 120 m high cliff. Assuming air resistance is negligible, how long is the ball in the air? What is the ball’s speed when it strikes the ground at the base of the cliff?t = 4.95svf =48.5m/sWhat’s your reaction time?Using a partner, vertical kinematics equations and a ruler, determine your reaction time…Average is 0.2s
19Free Fall – an object thrown upward The motion may be symmetricalThen tup = tdownThen the speed is the same at the same heights, but direction is oppositeThe motion may not be symmetricalBreak the motion into various partsGenerally up and down
20Symmetry in Free FallWhen something is thrown straight upward under the influence of gravity, and then returns to the thrower, this is very symmetrical = parabolaThe object spends half its time traveling up; half traveling down.Velocity when it returns to the ground is the opposite of the velocity it was thrown upward with.Acceleration is 9.8 m/s2 and directed DOWN the entire time the object is in the air!
21For symmetrical situations… Take each side separately…Only have to solve one side…Up SideInitial velocity is upward.a = g = m/s2 (slowing down)The instantaneous velocity at the maximum height is zeroFinal velocity is zero.Down SideInitial velocity is zero.a = g = 9.80 m/s2 (speeding up)The instantaneous velocity at the maximum height is zeroFinal velocity is downward.v = 0Time up = Time down
22Non-symmetrical Free Fall Need to divide the motion into segmentsPossibilities includeUpward and downward portionsThe symmetrical portion back to the release point and then the non-symmetrical portion