Kinematics Kinematic Equations and Falling Objects

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

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 = v 0 + at Note that Equation crosses the velocity ax graphical interpretatio I of Figure 2 9, the equ Also, note that 1-0.5 m Equation 2 7 has the sa v avg = v i + v f 2 v v EXERCISE 2 2 A ball is thrown straight of the ball is -9.81 m/s 2 a. 0.50 s, and b. Solution 1 v av = 2(v 0 + v) a. Substituting t = 0.5 (a) 1 Figure from James S. Walker Physics. v 0 O t t v = b. Similarly, using t = v =

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.

t will stop farther away from its starting point, so the answer to uation Using2.15 the that Kinematics if v Equations, Ex 2.8 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 hidden trooper. 1 Serway & Jewett, pg 39.

t will stop farther away from its starting point, so the answer to uation Using2.15 the that Kinematics if v Equations, Ex 2.8 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 conundeden Answer: Figure t = 31.0 2.13 s(example 2.8) A speeding car passes a hid- trooper. 1 Serway & Jewett, pg 39.

Summary the kinematic equations using kinematics equations Collected Homework Due Friday, Oct 6. Quiz Tomorrow at the start of class. (Uncollected) Homework Serway & Jewett, Ch 2, onward from page 49. Conceptual Q: 7; Problems: 25, 29, 33, 39, 83