 # Horizontal and Vertical Equations in one dimension

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Horizontal and Vertical Equations in one dimension
Kinematics Horizontal and Vertical Equations in one dimension

So far… Displacement → Δx = xf – xi Velocity → v = Δx
BASIC EQUATIONS LIST Displacement → Δx = xf – xi Velocity → v = Acceleration → a = Δx Δt Δv Δt

If 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 motion x t vf vi a If we know 3 of these, we can figure out the other 2.

From before we know: This equation can be rearranged to look like: This equation is the 1st kinematics equation.

If 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.

For 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 + 2ax Most of the time, xi will be 0.

Kinematics Formula Summary
For 1-D motion with constant acceleration: vf = vi + a t x = vi t + ½ a t 2 vf2 = vi2 + 2 a x

To Remember… If there is acceleration, we need more than just x = vt
Because the object has either increased speed, therefore going further Or the object has decreased speed, therefore not covering as much ground.

When 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 time Given 90mph 4.5cm We know that the ball is stopping

Example continued vf = vi + at x = vit + ½at2 vf2 = vi2 + 2ax
Looking for time = t Given 90mph = vi = 40.2m/s 4.5cm = x = 0.045m stopping = vf = 0 vf = vi + at x = vit + ½at2 vf2 = vi2 + 2ax Which equation do we use? Can we solve with just one? Need to find acceleration first. Use #3. a = m/s2 Then find time. Use #1. s or ms Answer

You 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 =161.3m t = 14.7 s a = -12m/s2 x = 37.5m

Galileo Galilei In addition to telescopes and his other pursuits Galileo formulated the laws that govern the motion of objects in free fall

Free Fall All objects moving under the influence of gravity only are said to be in free fall Free fall does not depend on the object’s original motion All objects falling near the earth’s surface fall with a constant acceleration Gravity 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”

Acceleration due to Gravity
g = 9.8 m/s² When estimating, use g » 10 m/s2 g is always directed downward Toward the center of the earth Ignoring air resistance and assuming g doesn’t vary with altitude over short vertical distances, free fall is constantly accelerated motion

Vertical Kinematics Equations
Replace a with g and x with y vf = vi + gt y = vit + ½gt2 vf2 = vi2 + 2gy

Free Fall of a dropped object
Initial velocity is zero Use the same general kinematics equations Generally use y instead of x since vertical Acceleration is g = 9.8m/s2 Because the object is speeding up, we will say that g is positive This does not agree with your book. vi = 0 a = g

Free Fall of an object thrown downward
a = g = 9.80 m/s2 Initial velocity  0 g is positive because the object is speeding up. We can choose that downward direction is positive when dealing with falling objects so Initial velocity will be positive The only thing acting on the object after launch is gravity… FREE FALL vi  0 a = g

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 = 4.95s vf =48.5m/s What’s your reaction time? Using a partner, vertical kinematics equations and a ruler, determine your reaction time… Average is 0.2s

Free Fall – an object thrown upward
The motion may be symmetrical Then tup = tdown Then the speed is the same at the same heights, but direction is opposite The motion may not be symmetrical Break the motion into various parts Generally up and down

Symmetry in Free Fall When something is thrown straight upward under the influence of gravity, and then returns to the thrower, this is very symmetrical = parabola The 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!

For symmetrical situations…
Take each side separately… Only have to solve one side… Up Side Initial velocity is upward. a = g = m/s2 (slowing down) The instantaneous velocity at the maximum height is zero Final velocity is zero. Down Side Initial velocity is zero. a = g = 9.80 m/s2 (speeding up) The instantaneous velocity at the maximum height is zero Final velocity is downward. v = 0 Time up = Time down

Non-symmetrical Free Fall
Need to divide the motion into segments Possibilities include Upward and downward portions The symmetrical portion back to the release point and then the non-symmetrical portion

Combination Motions

Multi-step Problems How fast should you throw a ball straight down from 40 m up so that its impact speed would be the same as a rock’s impact speed dropped from 60 m? Answer: 19.8 m/s