3. Displacement: - is the change in of an object. - it’s a term: i.e has & magnitude. not a term: i.e has magnitude only. Displacement is defined as: ∆x = x final – x initial ∆ (delta) means “ ” = x f – x i (path is ) +  forward movement -  backward movement x i x f 5m50m ∆x = x final – x initial = x f – x i =

4. x f x i 30m100m ∆x = x final – x initial = x f – x i = x i x f -30m0m50m ∆x = x f – x i = = 0 5m 10m50m ∆x = x f – x i = BUT: distance travelled = The difference is, this time, path is.

5. Velocity: Average velocity (v) is the rate of change of displacement (∆x) over a time interval. (∆t) velocity = displacementspeed = time vector v = dx = ∆x = x f - x i s = dt ∆t ∆t  Calculus !!! Ughh !

6. Lets look at a Displacement – Time graph. - A car ride to San Diego! What is the average velocity of the journey? v = ∆x = x f - x i = ∆t ∆t

7. Instantaneous Velocity: - is the velocity travelled at a given of time. - is the slope of the graph at that point in time. - remember Math – Aaahhh! - for a curved line, use as small a time limit as possible. i.e. Hint: If the point is part of a straight line segment, use as much of that segment as possible for more accurate results.

8. What is the velocity at 45 mins into the journey? v = lim ∆x = ∆t  0 ∆t What’s happening at the 105min (1.75hr) mark? What’s the signifigance of the angle of the slope?

9. 2 types:i) ii) acceleration is the change in velocity over a given period of time. a = ∆v = v f – v i ∆t t f – t i acceleration is the change in velocity at a given instant of time Inst.a = lim ∆v ∆t  0 ∆t

10. On a graph: Instantaneous acceleration is the at a particular point on the graph. Negative acceleration is also known as If y = f(x), then y’ = slope (acceleration). Sign (  ) tells direction a ∫ b y = area under curve =

11. A set of equations that show the relationship between displacement (x), velocity (v), acceleration (a) and time (t). They rely on 2 assumptions: i) the acceleration is ii) the motion is

12. 1) v f = v o + at(v o = v i ) 2) v = v o + v f = ½ (v o + v f ) 2 3) x = v.t = ½ (v o + v f ).t 4) x = v o t + ½ at 2 5) v f 2 = v o 2 + 2ax

14. Example: A car that is travelling at 40 m/s (? mph) brakes to a halt in 5 seconds. a.) Calculate the average acceleration (deceleration), and b.) the distance required to stop. Remember: Diagram & 5 steps! v o = 40 m/sa = ?v f = 0 t o = 0x = ?t f = 5s

16. A car at rest accelerates at a constant rate of 3 m/s 2. When the car has reached 25m, find: a) the velocity, and b) the time taken to reach that point. v 0 = 0a = 3m/s 2 v f = ? x = 25m ∆t = ?

18. A car at rest accelerates at 5 m/s 2 until it reaches 20 m/s. It then continues at constant velocity for 10 s. Calculate the total time and distance for this journey. Pt.1 |Pt.2 0m/s20m/s 20m/s 5m/s 2 10s

20. In the x direction, acceleration can vary depending on the object or other influences. In the y direction, as an object moves up or down, it is solely under the. More specifically – This is considered a – assuming it is not powered. Neglecting friction, ALL objects, regardless of size and mass, fall at the same rate of Therefore: a y = or more appropriately g =.

21. To go from the x direction to the y direction, we make a couple of substutions: xbecomes,and abecomes (why negative?) 1.)v yf = v yo - gt 4.)y = v yo t - ½ gt 2 5.)v yf 2 = v yo 2 - 2gy

22. Note:an object thrown vertically upwards with velocity (v), will come back with an e.g.Little Jonnie at the top of a 5 storey bldg (40m) drops a ball out the window. How long will it take to hit the ground and with what velocity? Note: -For falling objects, y has a value. -When objects hit/land on the ground, it’s velocity is 40m

24. E.g.2) Now little Jonnie throws a second ball upwards with a velocity of 25 m/s. Calculate the height above the ground the ball reaches, total time in the air, and the velocity at which it hits the ground. v y = 0 @ the apex (turning pt.) y T = height above the ground = t T = total time = or someway easier?

26. y = v yo t - ½ gt 2 because v yo  0, need to use the quadratic. ½ gt 2 – v yo t + y = 0 a bc a = ½ g = ½ (9.80m/s 2 ) = 4.90m/s 2 b = -v yo = -25m/s c = y = -40m t = -b   b 2 – 4ac 2a t = s, s so t =