1. Motion in One Dimension Unit 1

2. Lesson 1 : Position, Velocity, and Speed Position : location of a particle with respect to a chosen reference point Displacement : the change in position in some time interval  x = x f – x i Distance : the length of a path followed by a particle Displacement is a VECTOR QUANTITY Distance is a SCALAR QUANTITY

3. Average Velocity : a particle’s displacement (  x) divided by the time interval (  t) during which that displacement occurs V = xx tt Average Speed : the total distance traveled divided by the total time interval required to travel that distance Avg speed = total distance total time Avg Velocity is a VECTOR QUANTITY Avg Speed is a SCALAR QUANTITY

4. Example 1 Positiont(s)x(m) A030 B1052 C2038 D300 E40-37 F50-53 Find the displacement, average velocity, and average speed of the object between positions A and F.

5. Lesson 2 : Instantaneous Velocity and Speed Instantaneous Velocity : the limiting value of the ratio  x/  t as  t approaches zero v = xx tt lim  t  0 Instantaneous Speed : the magnitude of the instantaneous velocity

6. Example 1 A) A ball thrown directly upward rises to a highest point and falls back in the thrower’s hand. B) A race car starts from rest and speeds up to 100 m/s. C) A spacecraft drifts through space at constant velocity. Are there any points at which the instantaneous velocity has the same value as the average velocity over the entire motion ? If so, identify the point(s).

7. What is the instantaneous velocity at t = 5.0 s ? Answer : Slope of the tangent line drawn at the time in question. Example 2

8. Lesson 3 : Acceleration Average Acceleration : the change in velocity (  v) divided by the time interval (  t) during which that change occurs a = vv tt = v f – v i t f – t i Instantaneous Acceleration : the limit of the average acceleration as  t appoaches zero a = vv tt lim  t  0

9. Example 1 What is the instantaneous acceleration at t = 2.0 s ? Answer : Slope of the tangent line drawn at the time in question.

10. Example 2 The velocity of a particle moving along the x- axis varies in time according to the expression v x = (40 – 5t 2 ) m/s, where t is in seconds. A) Find the average acceleration in the time interval t =0 to t = 2.0 s. B) Determine the acceleration at t = 2.0 s.

11. Lesson 4 : Motion Diagrams v No acc Motion A v a Motion B v a Motion C

12. Graphs for each motion : x Motion A t v t a t x Motion B t v t a t x Motion C t v t a t

13. Example 1 Draw the corresponding x vs. t and a vs. t graphs. A) v t x t a t B) v t x t a t v C) t x t a t

14. Example 2 Draw the corresponding v vs. t and a vs. t graphs. A) x t v t a t B) x t v t a t x C) t v t a t

15. What is the displacement of the object from t = 0 to 3 s ? Answer : The area under the graph equals displacement Example 3

16. Negative Area : Object is moving toward smaller x values and displacement is decreasing

17. Lesson 5 : Kinematic Equations a = vv tt = v f – v i t f – t i v f = v i + a  t v f = v i + at (for constant a) v = v i + v f 2 (for constant a)

18.  x = x f - x i  x = vt x f – x i = vt x f – x i = ½ (v i + v f )t x f = x i + ½ (v i + v f )t (for constant a)

19. x f = x i + ½ (v i + v f )t x f = x i + ½ [v i + (v i +at)]t v f = v i + at x f = x i + v i t + ½ at 2 (for constant a) x f = x i + ½ (v i + v f )t x f = x i + ½ (v i + v f ) ( v f - v i a ) v f 2 = v i 2 + 2a (x f – x i ) (for constant a)

20. Summary v f = v i + at (for constant a) v = v i + v f 2 (for constant a) x f = x i + ½ (v i + v f )t (for constant a) x f = x i + v i t + ½ at 2 (for constant a) v f 2 = v i 2 + 2a (x f – x i ) (for constant a)

21. Example 1 A jet lands on an aircraft carrier at 140 mph (63 m/s). A)What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable that snags the airplane and brings it to a stop ? B)If the plane touches down at position x i = 0, what is the final position of the plane ?

22. Example 2 A car traveling at a constant speed of 45.0 m/s passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch it, accelerating at a constant rate of 3.00 m/s 2. How long does it take her to overtake the car ?

23. Lesson 6 : Freely Falling Objects Neglecting air resistance, all objects dropped near the Earth’s surface fall toward the Earth with the same constant acceleration under the influence of the Earth’s gravity. A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion.

24. Free-fall acceleration (g) = 9.80 m/s 2 +y-y a y = -g = -9.80 m/s 2 For making quick estimates, use g = 10 m/s 2

25. Example 1 A ball is tossed straight up at 25 m/s. Estimate its velocity at 1 s intervals. Time (s) Velocity (m/s) 0+25 1 2 3 4 5

26. Example 2 A stone is thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down. A)Determine the time at which the stone reaches its maximum height. B)Determine the maximum height.

27. D)Determine the velocity of the stone at this instant. E)Determine the velocity and position of the stone at t = 5.00 s. C)Determine the time at which the stone returns to the height from which it was thrown.

28. Lesson 7 : Using Calculus in Kinematics Instantaneous Velocity v = xx tt lim  t  0 This limit is called the derivative of x with respect to t. v = xx tt lim  t  0 = dx dt v = dx dt

29. f(x) = ax n Derivative of Power Function df dx = nax n-1 Example 1 An object is moving in one dimension according to the formula x(t) = 2t 3 + t 2 – 4. Find its velocity at t = 2 s.

30. Instantaneous Acceleration a = vv tt lim  t  0 This limit is called the derivative of v with respect to t. a = vv tt lim  t  0 = dv dt a = dv dt

31. Example 2 An object is moving in one dimension according to the formula v(t) = (40 – 5t 2 ). Find its acceleration at t = 2 s. Acceleration is also the second derivative of position. a = vv tt lim  t  0 = dv dt = d2xd2x dt 2

32. Example 3 An object is moving in one dimension according to the formula x(t) = 12 – 4t + 2t 3. Find its acceleration at t = 3 s.

33. The Integral or Antiderivative v t titi tftf tntn vnvn  x n = v n  t n = shaded area  x =  v n  t n = total area

34. As rectangles get narrower,  t n  0  x = lim  t n  0  v n  t n Displacement = area under v-t graph This is called the Definite Integral lim  t n  0  v n  t n = v(t) dt  titi tftf

35. Integrating Velocity v(t) = dx dt dx = v dt  xixi xfxf dx =  titi tftf v dtx f – x i =  titi tftf v dt

36. Integrating Acceleration a(t) = dv dt dv = a dt  vivi vfvf dv =  titi tftf a dtv f – v i =  titi tftf a dt

37. Finding Antiderivatives If k is a constant,  k dx = kx + C  x n dx = x n+1 n + 1 + C (n = 1)

38. Example 4 A car currently moving at 10. m/s accelerates non-uniformly according to a(t) = 3t 2. Find its velocity at t = 2 s.

39. Example 5 An object is moving according to the formula v(t) = 5t 2 – 4t. If the object starts from rest, find its position at t = 5s.

40. Example 6 An object is moving according to the formula a(t) = 2t – 4, with an initial velocity of + 4 m/s, and an initial position x = 0 at time t = 0. Find the position and velocity at arbitrary times.