Welcome back to Physics 215

1. Welcome back to Physics 215 Today’s agenda: Using graphs to solve problems Constant acceleration Free fall

2. Current homework assignment • HW1: • Ch.1 (Knight): 34, 40, 52, 56; Ch.2 (Knight): 34, 36, 40 • due Wednesday Sept 8th in recitation • TA will grade each HW set with a score from 0 to 5 Lecture 01-2

3. Workshops Two sections! • W-F 8:30-9:30AM and 4:30-5:30PM • Both meet in room 208 • Attend either section • Don’t need to change registration Labs • Sign up for whatever section of PHY 221 is compatible with your schedule and has space available. • *Office hours will end at 3:00PM today* Lecture 01-2

4. Interpreting x(t) and v(t) graphs x • Slope at any instant in x(t) graph gives instantaneous velocity • Slope at any instant in v(t) graph gives instantaneous acceleration • What else can we learn from an x(t) graph? t v t

5. Acceleration from x(t) • Rate of change of slope in x(t) plot equivalent to curvature of x(t) plot • Mathematically, we can write this as a = • Negative curvature • a < 0 • Positive curvature • a > 0 • No curvature • a = 0 x t

6. Sample problem • An object’s position as a function of time is given by x(t) = (3.00 m) - (2.00 m/s) t + (3.00 m/s2) t2. • Calculate the avg. accel. between t = 2.00s and t = 3.00 s. • Calculate the instantan. accel. at (i) t = 2.00 s; (ii) t = 3.00 s.

7. Displacement from velocity curve? v • Suppose we know v(t) (say as graph), can we learn anything about x(t) ? • Consider a small time interval Dt v = Dx/Dt  Dx = vDt t • So, total displacement is the sum of all these small displacements Dx • x = SDx = limDt0S v(t)Dt =

8. Graphical interpretation v v(t) T1 T2 t Dt Displacement between T1 and T2 is area under v(t) curve

9. Displacement – integral of velocity limDt0 S Dt v(t) = area under v(t) curve note: `area’ can be positive or negative *Consider v(t) curve for cart in different situations… v *Net displacement? t

10. Velocity from acceleration curve • Similarly, change in velocity in some time interval is just area enclosed between curve a(t) and t-axis in that interval. a a(t) t T1 T2 Dt

11. Summary • accel. a = dv/dt • = slope of v(t) curve • NOT v/t !! • change in vel.Dvis • ∫a(t)dt • = area under a(t) curve • NOT at !! • velocity v = dx/dt = slope of x(t) curve • NOT x/t !! • displacementDxis ∫v(t)dt = area under v(t) curve • NOT vt !!

12. a x b t T • The graph shows 2 trains running on • parallel tracks. Which is true: • At time T both trains have same v • Both trains speed up all time • Both trains have same v for some t<T • Somewhere, both trains have same a

13. Simplest case with non-zero acceleration • Constant acceleration: a = aav • Can find simple equations for x(t), v(t) in this case

14. 1st constant acceleration equation • From definition of aav: aav = Dv/Dt Let aav = a, Dt = t, Dv = v - v0 Find: v= v0 + at *equation of straight line in v(t) plot v v0 t

15. 2nd const. acceleration equation • Notice: graph makes it clear that vav=(1/2)(v + v0) x - x0 = (1/2)(v + v0)t v v0 t

16. An object moves with constant acceleration for 40 seconds, covering a total distance of 800 meters. Its average velocity is therefore 20 m/s. Its instantaneous velocity will be equal to 20 m/s… 1. at t = 20 seconds (i.e., half the time) 2. at s = 400 meters (i.e., half the distance) 3. both of the above. 4. Cannot tell based on the information given.

17. 3rd constant acceleration equation • Using 1st constant acceleration equation v= v0 + at insert into relation between x, t, and vav : x - x0 = vavt = (1/2)(v + v0)t (set Dt = t, i.e. take t0 = 0) yields: x - x0 = (1/2)(2v0 + at)t or: x = x0 + v0t + (1/2)at2

18. x(t) graph- constant acceleration x = x0 + v0t + (1/2)at2 x parabola *Initial sign of v ? *Sign of a ? t

19. 4th constant acceleration equation • Can also get an equation independent of t • Substitute t = (v - v0)/a into x - x0 = (1/2)(v+ v0)t we get: 2a(x - x0) = v2 - v02 or: v2 = v02 + 2a(x - x0)

20. Example of constant acceleration: • Free fall demo • Compare time taken by feather and billiard ball to fall to the ground from the same height • Influence of air in room?

21. Freely Falling Bodies: • Near the surface of the earth, neglecting air resistance and the earth’s rotation, all objects experience the same downward acceleration due to their gravitational attraction with the earth: • Near = height small relative to radius of earth g = 9.8 m/s2

22. Motion with constant acceleration: v= v0 + at vav= (1/2) (v0+ v) x= x0 + v0t + (1/2) at2 v2= v02 + 2a (x - x0) *where x0, v0 refer to time = 0 s ; x, v to time t

23. An object moves with constant acceleration, starting from rest at t = 0 s. In the first four seconds, it travels 10 cm. What will be the displacement of the object in the following four seconds (i.e. between t = 4 s and t = 8 s)? 1. 10 cm 2. 20 cm 3. 30 cm 4. 40 cm

24. Rolling disk demo • Compute average velocity for each section of motion (between marks) • Measure time taken (metronome) • Compare v at different times (i) 10 cm (ii) 30 cm (iii) 50 cm (iv) 70 cm v t

25. Sample problem • A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 2.50 s. Ignore air resistance, so the brick is in free fall. • How tall, in meters, is the building? • What is the magnitude of the brick’s velocity just before it reaches the ground? • Sketch a(t), v(t), and y(t) for the motion of the brick.

26. Reading assignment • Vectors, 2D motion • 3.1-3.4, and reread 1.3 in textbook