Kinematics Kinematic Equations and Falling Objects Lana Sheridan De Anza College Sept 28, 2017
Last time kinematic quantities relating graphs
Overview derivation of kinematics equations using kinematics equations falling objects and g
Equations that describe motion in terms of position, velocity, acceleration, and time. To solve kinematics problems: identify what equations apply, then do some algebra to find the quantity you need.
For constant velocity (a = 0): x = vt (1)
For constant velocity (a = 0): x = vt (1) In general, when velocity is not constant, this is always true: x = v avg t
Always true: x = v avg t (2) This is just a rearrangement of the definition of average velocity: average velocity, v avg v avg = x t where x is the displacement that occurs in a time interval t.
If acceleration is constant (a = const): From v avg = x t, we also have: v avg = v i + v f 2
Recall, x = v avg t and v avg = v i+v f 2 (constant acceleration) Substituting the second in to the first: ( vi + v f x = 2 ) t (3)
From the definition of acceleration: We can integrate this expression. a = dv dt
From the definition of acceleration: We can integrate this expression. For constant acceleration: a = dv dt v(t) = v i + at (4) where v i is the velocity at t = 0 and v(t) is the velocity at time t.
For constant acceleration: x(t) = x i + v i t + 1 2 at2 Equivalently, x = v i t + 1 2 at2 (5) How do we know this?
For constant acceleration: x(t) = x i + v i t + 1 2 at2 Equivalently, x = v i t + 1 2 at2 (5) How do we know this? Integrate!
The last equation we will derive is a scaler equation.
The last equation we will derive is a scaler equation. ( ) vi + v f x = t 2 We could also write this as: ( vi + v f ( x) i = 2 ) t i where x, v i, and v f could each be positive or negative.
The last equation we will derive is a scaler equation. ( ) vi + v f x = t 2 We could also write this as: ( vi + v f ( x) i = 2 ) t i where x, v i, and v f could each be positive or negative. We do the same for equation (4): v f i = (v i + at) i
The last equation we will derive is a scaler equation. ( ) vi + v f x = t 2 We could also write this as: ( vi + v f ( x) = 2 ) t where x, v i, and v f could each be positive or negative. We do the same for equation (4): v f = (v i + at) Rearranging for t: t = v f v i a
t = v f v i a ( vi + v f ; x = 2 Substituting for t in our x equation: x = ( ) ( ) vi + v f vf v i 2 a 2a x = (v i + v f )(v f v i ) so, ) t v 2 f = v 2 i + 2 a x (6)
For zero acceleration: Always: x = vt For constant acceleration: x = v avg t v f = v i + at x = v i t + 1 2 at2 x = v i + v f t 2 vf 2 = vi 2 + 2 a x
Summary For zero acceleration: x = vt Always: x = v avg t For constant acceleration: v f = v i + at x = v i t + 1 2 at2 x = v i + v f t 2 vf 2 = vi 2 + 2 a x
Using the Kinematics Equations Example Drag racers can have accelerations as high as 26.0 m/s 2. Starting from rest (v i = 0) with that acceleration, how much distance does the car cover in 5.56 s? Before we start answering...
Using the Kinematics Equations Process: 1 Identify which quantity we need to find and which ones we are given. 2 Is there a quantity that we are not given and are not asked for? 1 If so, use the equation that does not include that quantity. 2 If there is not, more that one kinematics equation may be required or there may be several equivalent approaches. 3 Input known quantities and solve.
Using the Kinematics Equations Example Drag racers can have accelerations as high as 26.0 m/s 2. Starting from rest (v i = 0) with that acceleration, how much distance does the car cover in 5.56 s? 1 OpenStax Physics
t will stop farther away from its starting point, so the answer to Using the Kinematics Equations, Ex 2.8 uation 2.15 that if v xi is larger, then x f will be larger. imit! A car traveling at a constant speed of 45.0 m/s passes a trooper on a motorcycle hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch the car, accelerating at a constant rate of 3.00 AM m/s 2. How long does it take the trooper to overtake the car? sses a e secooper g at a ooper t 1.00 s t 0 t? y the er conunder Figure 2.13 (Example 2.8) A speeding car passes a hid- 1 Serway den trooper. & Jewett, pg 39.
Falling Objects One common scenario of interest where acceleration is constant is falling objects. Objects (with mass) near the Earth s surface accelerate at a constant rate of g = 9.8 ms 2. (Or, about 10 ms 2 ) The kinematics equations for constant acceleration all apply.
Falling Objects One common scenario of interest where acceleration is constant is falling objects. Objects (with mass) near the Earth s surface accelerate at a constant rate of g = 9.8 ms 2. (Or, about 10 ms 2 ) The kinematics equations for constant acceleration all apply. Example You drop a rock off of a 50 m tall building. With what velocity does it strike the ground? How long does it take to fall to the ground?
Falling Objects CHAPTER 2 KINEMATICS 65 Figure 2.40 Vertical position, vertical velocity, and vertical acceleration vs. time for a rock thrown vertically up at the edge of a cliff. Notice that velocity changes linearly with time and that acceleration is constant. Misconception Alert! Notice that the position vs. time graph shows vertical position only. It is easy to get the impression that the graph shows some horizontal motion the shape of the graph looks like the path of a projectile. But this is not the case; the horizontal axis is time, not space. The 1 OpenStax Physics
Acceleration in terms of g Because g is a constant, and because we have a good intuition for it, we can use it as a unit of acceleration. ( How many times g is this acceleration? )
Acceleration in terms of g Because g is a constant, and because we have a good intuition for it, we can use it as a unit of acceleration. ( How many times g is this acceleration? ) Consider the drag race car in the earlier example. For the car the maximum acceleration was a = 26.0 m s 2. How many gs is that? A 0 B roughly 1 C roughly 2 and a half D 26
Acceleration of a Falling Object Question A baseball is thrown straight up. It reaches a peak height of 15 m, measured from the ground, in a time 1.7 s. Treating up as the positive direction, what is the acceleration of the ball when it reached its peak height? A 0 m/s 2 B 8.8 m/s 2 C +8.8 m/s 2 D 9.8 m/s 2 1 Princeton Review: Cracking the AP Physics Exam
Summary the kinematic equations using kinematics equations falling objects Collected Homework Due Friday, Oct 6. (Uncollected) Homework Serway & Jewett, Ch 2, onward from page 49. Conceptual Q: 7; Problems: 25, 29, 33, 39, 83