1. Physics 215 – Fall 2014Lecture 02-11 Welcome back to Physics 215 Today’s agenda: Using graphs to solve problems Constant acceleration Free fall

2. Physics 215 – Fall 2014Lecture 02-12 2 Current homework assignment HW1: –Ch.1 (Knight): 42, 52, 56; Ch.2 (Knight): 26, 30, 40 –due Wednesday, Sept 3 rd in recitation –TA will grade each HW set with a score from 0 to 10

3. Physics 215 – Fall 2014Lecture 02-13 3 Workshops Two sections! –Wednesdays: 8:25-9:20AM and 2:35-3:30PM –Fridays: 8:25-9:20AM and 10:35-11:30AM –Both meet in room 208 –Attend either section each day –Don’t need to change registration

4. Physics 215 – Fall 2014Lecture 02-14 4 Acceleration Similarly when velocity changes it is useful (crucial!) to introduce acceleration a a av =  v/  t = (v F - v I )/  t Average acceleration -- keep time interval  t non-zero Instantaneous acceleration a inst = lim  t  0  v/  t = dv/dt

5. Physics 215 – Fall 2014Lecture 02-15 5 Sample problem A car’s velocity as a function of time is given by v(t) = (3.00 m/s) + (0.100 m/s 3 ) t 2. –Calculate the avg. accel. for the time interval t = 0 to t = 5.00 s. –Calculate the instantaneous acceleration for i) t = 0; ii) t = 5.00 s.

6. Physics 215 – Fall 2014Lecture 02-16 6 Acceleration from graph of v(t) t v P Q T What is a av for PQ ? QR ? RT ? R Slope measures acceleration –Positive a means v is increasing –Negative a means v decreasing

7. Physics 215 – Fall 2014Lecture 02-17 Interpreting x(t) and v(t) graphs 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? x t v t

8. Physics 215 – Fall 2014Lecture 02-18 t x b a The graph shows 2 trains running on parallel tracks. Which is true: 1.At time T both trains have same v 2.Both trains speed up all time 3.Both trains have same v for some t<T 4.Somewhere, both trains have same a T

9. Physics 215 – Fall 2014Lecture 02-19 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

10. Physics 215 – Fall 2014Lecture 02-110 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/s 2 ) t 2. –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.

11. Physics 215 – Fall 2014Lecture 02-111 Displacement from velocity curve? Suppose we know v(t) (say as graph), can we learn anything about x(t) ? Consider a small time interval  t v =  x/  t   x = v  t So, total displacement is the sum of all these small displacements  x x =   x = lim  t  0  v(t)  t = v t

12. Physics 215 – Fall 2014Lecture 02-112 Graphical interpretation v t TT T2T2 tt v(t) Displacement between T 1 and T 2 is area under v(t) curve

13. Physics 215 – Fall 2014Lecture 02-113 Displacement – integral of velocity lim  t  0  t v(t) = area under v(t) curve note: `area’ can be positive or negative *Consider v(t) curve for cart in different situations… v t *Net displacement?

14. Physics 215 – Fall 2014Lecture 02-114 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 t TT T2T2 tt a(t)

15. Physics 215 – Fall 2014Lecture 02-115 Summary velocity v = dx /dt = slope of x(t) curve –NOT x/t !! displacement  x  is ∫ v(t)dt = area under v(t) curve –NOT vt !! accel. a = dv /dt = slope of v(t) curve –NOT v/t !! change in vel.  v  is ∫ a(t)dt = area under a(t) curve –NOT at !!

16. Physics 215 – Fall 2014Lecture 02-116 Simplest case with non-zero acceleration Constant acceleration: a = a av Can find simple equations for x(t), v(t) in this case

17. Physics 215 – Fall 2014Lecture 02-117 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?

18. Physics 215 – Fall 2014Lecture 02-118 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/s 2

19. Physics 215 – Fall 2014Lecture 02-119 1 st constant acceleration equation From definition of a av :a av =  v/  t Let a av = a,  t = t,  v = v - v 0 Find: v = v 0 + at *equation of straight line in v(t) plot t v v0v0

20. Physics 215 – Fall 2014Lecture 02-120 2 nd const. acceleration equation Notice: graph makes it clear that v av =(1/2)(v + v 0 ) x - x 0 = (1/2)(v + v 0 )t t v v0v0

21. Physics 215 – Fall 2014Lecture 02-121 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.

22. Physics 215 – Fall 2014Lecture 02-122 3 rd constant acceleration equation Using 1 st constant acceleration equation v = v 0 + at insert into relation between x, t, and v av : x - x 0 = v av t = (1/2)(v + v 0 )t (set  t = t, i.e. take t 0 = 0) yields: x - x 0 = (1/2)(2v 0 + at)t or: x = x 0 + v 0 t + (1/2)at 2

23. Physics 215 – Fall 2014Lecture 02-123 x(t) graph- constant acceleration x t parabola *Initial sign of v ? *Sign of a ? x = x 0 + v 0 t + (1/2)at 2

24. Physics 215 – Fall 2014Lecture 02-124 4 th constant acceleration equation Can also get an equation independent of t Substitute t = (v - v 0 )/a into x - x 0 = (1/2)(v + v 0 )t we get: 2a(x - x 0 ) = v 2 - v 0 2 or: v 2 = v 0 2 + 2a(x - x 0 )

25. Physics 215 – Fall 2014Lecture 02-125 Motion with constant acceleration: v  = v 0 + at v av  = ( 1 / 2 ) (v 0 + v) x = x 0 + v 0 t + ( 1 / 2 ) a t 2 v 2 = v 0 2 + 2a (x - x 0 ) *where x 0, v 0 refer to time = 0 s ; x, v to time t

26. Physics 215 – Fall 2014Lecture 02-126 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

27. Physics 215 – Fall 2014Lecture 02-127 Rolling disk demo Compute average velocity for each section of motion (between marks) Measure time taken (metronome) Compare v at different times v t (i) 10 cm (ii) 30 cm (iii) 50 cm (iv) 70 cm

28. Physics 215 – Fall 2014Lecture 02-128 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.

29. Physics 215 – Fall 2014Lecture 02-129 Motion in more than 1 dimension Have seen that 1D kinematics is written in terms of quantities with a magnitude and a sign Examples of 1D vectors To extend to d > 1, we need a more general definition of vector

30. Physics 215 – Fall 2014Lecture 02-130 Reading assignment Vectors, 2D motion 3.1-3.4, and reread 1.3 in textbook