1. Clickers Setup Turn on your clicker (press the power button) Set the frequency: –Press and hold the power button –Two letters will be flashing –If it’s not “AC”, press “A” and then “C” If everything works, you should see “Welcome” and “Ready”

2. Kinematics of 1-dimensional motion Displacement: Average velocity: Instantaneous velocity: Average acceleration: Instantaneous acceleration: Curvature

3. Equations of Kinematics for Constant Acceleration Proof: Result: Position x, velocity v x, acceleration a x, and time t are related by the kinematic equations. There are two initial conditions x 0 and v 0x. Parabola

4. Freely Falling Bodies Acceleration due to gravity g=9.8 m/s 2 Example 2.6: Find velocity and position of a coin freely falling from rest after 1, 2, 3 s. Solution: 1.Choose axes and plot trajectory. 2.Indicate known and unknown data y0y0 v 0y yvyvy ayay t 0 0??-g=-9.8m/s 2 1, 2, 3 s By choice of axis Implied data 3. Write and solve equations in symbols 4. Do numerical calculations. 5. Check the answer: dimensions, signs, functional dependencies, scales…

5. Problem Solving Steps 1. Geometry & drawing: trajectory, vectors, coordinate axes free-body diagram, … 2. Data: a table of known and unknown quantities, including “implied data”. 3. Equations ( with reasoning comments ! ), their solution in algebraic form, and the final answers in algebraic form !!! 4. Numerical calculations and answers. 5. Check: dimensional, functional, scale, sign, … analysis of the answers and solution.

6. Exam Example 1 : Coin Toss V y =0 y 0 y0y0 v 0y yvyvy a y t 0+6m/s??-g=-9.8m/s 2 ? Questions: (a) How high does the coin go? (b) What is the total time the coin is in the air? Total time T= 2 t = 1.2 s (c) What is its velocity when it comes back at y=0 ? for y=0 and v y <0 yields v y 2 = v 0 2 → v y = -v 0 = - 6m/s (problem 2.85)

7. Position and Velocity by Integration Constant-acceleration formulas via integration

8. Example of kinematic problem with two possible solutions x0x0 v 0 xvxvx a x t 0+3250m/s215 km?-10m/s 2 ? Accelerating spacecraft data Question: Find velocity at a displacement x=215 km ? X=215 km = 215 000 m x

9. Two segments of motion with different accelerations x 01 v 01 x 1 v1v1 a 1 t1t1 00+120 m?+2.6 m/s 2 ? Segment 1 Segment 2 x 02 v 02 x2x2 v 2 a 2 t2t2 x1x1 v1v1 ?+12m/s-1.5 m/s 2 ? Question: Find total displacement D ? x 0 x 1 =x 02 Segment 1 Segment 2

10. Exam Example 2: Accelerated Car (problems 2.7 and 2.17) Data: x(t)= αt+βt 2 +γt 3, α=6m/s, β=1m/s 2, γ = -2 m/s 3, t=1s Find: (a) average and instantaneous velocities; (b) average and instantaneous accelerations; (c) a moment of time t s when the car stops. Solution: (a) v(t)=dx/dt= α+2βt +3γt 2 ; v 0 =α; (b) a(t)=dv/dt= 2β +6γt; a 0 =2β; (c) v(t s )=0 → α+2βt s +3γt s 2 =0 x 0 V(t) t 0 tsts α a(t) 2β2β

11. Exam Example 3: Truck vs. Car (problem 2.34) x 0 Data: Truck v=+20 m/s Car v 0 =0, a c =+3.2 m/s 2 Questions: (a) x where car overtakes the truck; (b) velocity of the car V c at that x; (c) x(t) graphs for both vehicles; (d) v(t) graphs for both vehicles. Solution: truck’s position x=vt, car’s position x c =a c t 2 /2 (a) x=x c when vt=a c t 2 /2 → t=2v/a c → x=2v 2 /a c (b) v c =v 0 +a c t → v c =2v x t 0 truck car truck car t V(t) v 0 t=2v/a c v/a c v c =2v t=2v/a c

12. Exam Example 4: Free fall past window (problem 2.84) Data: Δt=0.42 s ↔ h=y 1 -y 2 =1.9 m, v 0y =0, a y = - g Find: (a) y 1 ; (b) v 1y ; (c) v 2y y 0 y1y1 y2y2 V 0y =0 V 1y V 2y ayay h 1 st solution: (b) Eq.(3) y 2 =y 1 +v 1y Δt – gΔt 2 /2 → v 1y = -h/Δt + gΔt/2 (a) Eq.(4) → v 1y 2 = -2gy 1 → y 1 = - v 1y 2 /2g = -h 2 /[2g(Δt)] 2 +h/2 – g(Δt) 2 /8 (c) Eq.(4) v 2y 2 = v 1y 2 +2gh = (h/Δt + gΔt/2) 2 2 nd solution: (a)Free fall time from Eq.(3): t 1 =(2|y 1 |/g) 1/2, t 2 =(2|y 2 |/g) 1/2 → Δt+t 1 =t 2 (b) Eq.(4) → (c) Eq.(4) →

13. Exam Example 5: Relative motion of free falling balls (problem 2.94) y 0 H Data: v 0 =1 m/s, H= 10 m, a y = - g Find: (a) Time of collision t; (b) Position of collision y; (c) What should be H in order v 1 (t)=0. 1 2 Solution: (a) Relative velocity of the balls is v 0 for they have the same acceleration a y = –g → t = H/v 0 (b) Eq.(3) for 2 nd ball yields y = H – (1/2)gt 2 = H – gH 2 /(2v 0 2 ) (c) Eq.(1) for 1 st ball yields v 1 = v 0 – gt = v 0 – gH/v 0, hence, for v 1 =0 we find H = v 0 2 /g

14. Pulsars – Rotating Neutron Stars Discovery (1967) : “Little green men” Radio, optical, and X-ray pulses t Period T~ 1ms – 4s is astonishingly stable !!! T Pulse duration 10 km n, p +,e - Compare: R sun ~ 700 000 km Photons Neutron star Earth

15. Pulsar maps have been included on the two Pioneer Plaques as well as the Voyager Golden Record. They show the position of the Sun, relative to 14 pulsars, so that our position both in space and in time can be calculated by potential extraterrestrial intelligences. Pulsar positioning could create a spacecraft navigation system independently, or be an auxiliary device to GPS instruments.Pioneer Plaques Voyager Golden RecordSunextraterrestrial