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© 2010 Pearson Education, Inc. Motion Is Relative Displacement vs. Distance Velocity vs. Speed Velocity: Average and Instantaneous Acceleration Free Fall.

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Presentation on theme: "© 2010 Pearson Education, Inc. Motion Is Relative Displacement vs. Distance Velocity vs. Speed Velocity: Average and Instantaneous Acceleration Free Fall."— Presentation transcript:

1 © 2010 Pearson Education, Inc. Motion Is Relative Displacement vs. Distance Velocity vs. Speed Velocity: Average and Instantaneous Acceleration Free Fall I recommend writing down items in these yellow boxes Chapter 2: Motion in One Direction

2 © 2010 Pearson Education, Inc. Motion Is Relative – What does this mean? Motion of objects is always described as relative to something else. For example: Motion takes place over time and depends upon the frame of reference (precise location of objects in space).

3 © 2010 Pearson Education, Inc. Displacement – (your thoughts?) The length of the straight line drawn from its initial position to the objects final position. Displacement: the change in position of an object Δx = x f – x i Δ = Change x f = final position (on x-axis) x i (x o ) = initial position (on x-axis) Δx = Change in position on the x-axis

4 © 2010 Pearson Education, Inc. Displacement Δx = x f – x i Δx = Change in position on the x-axis xfxf xixi Δx = x f – x i = 7.0 cm – 2.0 cm = 5.0 cm

5 © 2010 Pearson Education, Inc. Displacement Δx = x f – x i Δx = Change in position on the x-axis xfxf xixi Δx = x f – x i = 1.2 cm – 6.5 cm = -5.3cm Positive or negative depends on the direction

6 © 2010 Pearson Education, Inc. Displacement Δy = y f – y i Δy = Change in position on the y-axis yfyf yiyi Δy = y f – y i = 10.0cm – 6.1 cm = 3.9cm Positive or negative depends on the direction

7 © 2010 Pearson Education, Inc. Displacement = Distance ??? xfxf xixi xfxf xixi

8 © 2010 Pearson Education, Inc. Displacement = Distance ??? xfxf xixi xfxf xixi xfxf Distance does not have positive or negative, it is the total distance traveled, NOT just the final outcome (x f – x i ).

9 © 2010 Pearson Education, Inc. Velocity – (your thoughts?) NOTE: For most problems, the initial values are going to be zero. We typically start at time zero, then begin. Our displacement also typically starts at a position of zero and moves a certain distance. Δx change in distance Δt change in time v avg = = = m/s Δx x f – x i Δt t f – t i v avg = = dtdt v = = m/s

10 © 2010 Pearson Education, Inc. Velocity = Speed ??? (your thoughts?) Average Velocity Constant Velocity Instantaneous Velocity

11 © 2010 Pearson Education, Inc. Velocity (v) is a description of –the instantaneous speed of the object –what direction the object is moving (north, left, right, up, down, pos x-axis, neg x-axis) Vectors have magnitude and direction Velocity is a vector quantity. It has –magnitude: instantaneous speed –direction: direction of objects motion (typically positive and negative x-axis) Δx change in distance Δt change in time v = = = m/s dtdt v = = m/s

12 © 2010 Pearson Education, Inc. Speed (s) and Velocity (v) speedConstant speed is steady speed, neither speeding up nor slowing down. velocityConstant velocity is –constant speed and –constant direction (straight-line path with no acceleration). Motion is relative to Earth, unless otherwise stated.

13 © 2010 Pearson Education, Inc. Example: A flying disc leaves Mr. Pearsons hand with a velocity of 15 m/s. Ignoring wind resistance, the disc travels for 10 seconds before hitting the ground. How far did the disc travel? Δd change in distance Δt change in time v = = = m/s dtdt v = = m/s v = 15 m/s t = 10 s d = ____ m dtdt v = d = t * v = 10 s * 15m/s d = 150 m Ignoring wind resistance = no friction = constant velocity P.I.E.S.S ??? v = 15 m/s

14 © 2010 Pearson Education, Inc. Velocity Δx x f – x i Δt t f – t i v avg = = t0t0 t1t1 t2t2 t3t3 x 0 = 0.0 cm x 1 = 1.5 cm x 2 = 4 cmx 3 = 9 cm 1) What is the displacement between x 1 and x 2 ? 2) What is the average velocity over the entire movement from x 0 to x 3 ? 3) What is the average velocity between t 0 and t 1 ? 4) What is the average velocity between x 2 and x 3 ?

15 © 2010 Pearson Education, Inc. ΔxΔx ΔtΔt time (s) ΔxΔt ΔxΔt Δx x f – x i Δt t f – t i v avg = = distance (m)

16 © 2010 Pearson Education, Inc. time (s) distance (m) Δx x f – x i Δt t f – t i v avg = = Instantaneous Velocity (speedometer)

17 © 2010 Pearson Education, Inc.

18 Instantaneous Velocity Instantaneous Velocity - is the velocity at any specific instant. Example: –When you ride in your car, you may speed up and slow down - (this will affect your average velocity over a given distance and time) –Your instantaneous speed is given by your speedometer (your speed at that moment).

19 © 2010 Pearson Education, Inc. Average Speed The entire distance covered divided by the total travel time –Doesnt indicate various instantaneous speeds along the way. Example: What is your average speed if you drive a total distance of 200 km in 2 hours? 200 km 2 h = 100 km/h

20 © 2010 Pearson Education, Inc. The average speed of driving 30 km in 1 hour is the same as the average speed of driving A.30 km in 1/2 hour. B.30 km in 2 hours. C.60 km in 1/2 hour. D.60 km in 2 hours. Average Speed CHECK YOUR NEIGHBOR

21 © 2010 Pearson Education, Inc. The average speed of driving 30 km in 1 hour is the same as the average speed of driving A.30 km in 1/2 hour. B.30 km in 2 hours. C.60 km in 1/2 hour. D.60 km in 2 hours. Average Speed CHECK YOUR ANSWER Explanation: Average speed = total distance / time So, average speed = 30 km / 1 h = 30 km/h. Now, if we drive 60 km in 2 hours: Average speed = 60 km / 2 h = 30 km/h Same 60 km 2 hour = 30 km 1 hour

22 © 2010 Pearson Education, Inc. Acceleration Formulated by Galileo based on his experiments with inclined planes.

23 © 2010 Pearson Education, Inc. Acceleration (a) - rate at which velocity changes over time Δv v f - v o change in velocity m/s Δt t f - t o time required for change s a = = = = = m/s 2 v f = velocity final v o = velocity initial (if starting from rest v o = ZERO) t f = time final t o = time initial NOTE: Usually we dont use t f - t o just t = required time a = = = m/s 2 v m/s t s v o = velocity at the starting time At time = 0 sec

24 © 2010 Pearson Education, Inc. Example: You cars speed right now is 40 km/h. ( v o = 40 km/h) Your cars speed 2 s later is 45 km/h. ( v f = 45 km/h t = 2s ) Your cars change in speed is 45 – 40 = 5 km/h. (change = Δv) Δv v f - v o change in velocity m/s Δt t time required for change s a = = = = = m/s 2 Δv v f - v o 45 km/h – 40 km/h 5 km/h Δt t 2 s 2 s a = = = = = 2.5 km/h/s Your cars acceleration is 2.5 km/h/s.

25 © 2010 Pearson Education, Inc. Δv v f - v o change in velocity m/s Δt t change in time s a = = = = = m/s 2 a = = = m/s 2 v m/s t s Example: A flying disc leaves Mr. Pearsons hand with a velocity of 15 m/s. After 3 seconds, wind resistance slows down the disc to 11 m/s. What is the acceleration of the disc?, v o = 15 m/s t = 3 s v f - v o t a = a = 1.3 m/s 2 Change in velocity results in Acceleration v f = 11 m/s 11 m/s – 15 m/s 3 s = -4 m/s 3 s = The negative means that the acceleration is in the direction opposite to the motion of the object = resulting in slowing

26 © 2010 Pearson Education, Inc. Δv v f - v o change in velocity m/s Δt t change in time s a = = = = = m/s 2 a = = = m/s 2 v m/s t s Example: If a different disc has an acceleration of -2.5 m/s 2. What will the velocity be after 2 seconds of flight if Mr. Pearson releases the disc at 13 m/s. v f = 8 m/s Dont wait for the teacher, get to work x Is this answer REASONABLE? Explain

27 © 2010 Pearson Education, Inc. Acceleration Acceleration involves a change in velocity: A change in speed (faster or slower), or A change in direction, or Both. When you are driving a car, what are three things you can do to cause acceleration?. NOTE: You do not feel velocity You can feel a CHANGE in velocity You DO feel ACCELERATION

28 © 2010 Pearson Education, Inc. An automobile is accelerating when it is A.slowing down to a stop. B.rounding a curve at a steady speed. C.Both of the above. D.Neither of the above. Acceleration CHECK YOUR NEIGHBOR

29 © 2010 Pearson Education, Inc. An automobile is accelerating when it is A.slowing down to a stop. B.rounding a curve at a steady speed. C.Both of the above. D.Neither of the above. Acceleration CHECK YOUR ANSWER Explanation: Change in speed (increase or decrease) is acceleration, so slowing is acceleration. Change in direction is acceleration (even if speed stays the same), so rounding a curve is acceleration.

30 © 2010 Pearson Education, Inc. Acceleration and velocity are actually A.the same. B.rates but for different quantities. C.the same when direction is not a factor. D.the same when an object is freely falling. Acceleration CHECK YOUR NEIGHBOR

31 © 2010 Pearson Education, Inc. Acceleration and velocity are actually A.the same. B.rates but for different quantities. C.the same when direction is not a factor. D.the same when an object is freely falling. Acceleration CHECK YOUR ANSWER Explanation: Velocity is the rate at which distance changes over time, Acceleration is the rate at which velocity changes over time. Δv v f - v o change in velocity m/s Δt t change in time s a = = = = = m/s 2 Δd change in distance Δt change in time v = = = m/s

32 © 2010 Pearson Education, Inc. Acceleration Galileo increased the inclination of inclined planes. Steeper inclines gave greater accelerations. When the incline was vertical, acceleration was max, same as that of the falling object. When air resistance was negligible, all objects fell with the same unchanging acceleration. Unchanging = CONSTANT Acceleration Which ball hits the ground first? Which ball has the highest v f ? Which ball has the highest v o ?

33 © 2010 Pearson Education, Inc. Free Fall Falling under the influence of gravity only - with no air resistance Freely falling objects on Earth accelerate at the rate of 10 m/s/s, i.e., 10 m/s 2 a = 10 m/s 2 = 9.8 m/s 2 = 9.81 m/s 2 (sig figs) We say this is the acceleration due to gravity (g): a = g = 10 m/s 2 = 9.8 m/s 2 Text book vs Real World (easier math)

34 © 2010 Pearson Education, Inc. Free FallHow Fast? The velocity acquired by an object starting from rest is So, under free fall, when acceleration is a = g = 10 m/s 2, the speed is v 0 = 0 m/s before release: t 0 = 0 s v 1 = 10 m/s after 1 s t 1 = 1 s v 2 = 20 m/s after 2 s. t 2 = 2 s v 3 = 30 m/s after 3 s. t 3 = 3 s And so on. a = ΔvtΔvt Acceleration = Change of 10 m/s per second

35 © 2010 Pearson Education, Inc. Gravity WITH Friction In this situation g still equals 10 m/s 2, but the motion of the object is no longer strait down. Here, we must include friction and trigonometry. In this example: a = 4 m/s 2 NOT 10 m/s 2 v 0 = 0 m/s before release: t 0 = 0 s v 1 = 4 m/s after 1 s t 1 = 1 s v 2 = 8 m/s after 2 s. t 2 = 2 s v 3 = 12 m/s after 3 s. t 3 = 3 s And so on. a = ΔvtΔvt Acceleration = Change of 4 m/s per second

36 © 2010 Pearson Education, Inc. A free-falling object has a speed of 30 m/s at one instant. Exactly 1 s later its speed will be A.the same. B.35 m/s. C.more than 35 m/s D.60 m/s. Free FallHow Fast? a = g = 10 m/s 2 a = ΔvtΔvt Δ v = a*t v f = v o + a*t v f = v o + Δv v f = 30m/s + (10m/s 2 * 1s) v f = 30m/s + 10m/s v f = 40m/s v o = 30 m/s t = 1 s v f = ____ m/s Freely Falling

37 © 2010 Pearson Education, Inc. Free FallHow Far? The distance covered by an accelerating object -- starting from rest is: So, under free fall, when acceleration is 10 m/s 2, the distance is at t = 1 s; d = ½ *a*(t) 2 = ½ (10 m/s 2 ) (1 s) 2 = 5 m at t = 2 s; d = ½ *a*(t) 2 = ½ (10 m/s 2 ) (2 s) 2 = 20 m at t = 3 s; d = ½ *a*(t) 2 = ½ (10 m/s 2 ) (3 s) 2 = 45 m And so on d = ½ *a*(t) 2 m = ½ *(m/s 2 )*(s) 2 Note that the seconds cancel

38 © 2010 Pearson Education, Inc. What is the distance covered of a freely falling object starting from rest after 4 s? A.4 m B.16 m C.40 m D.80 m Free FallHow Far? CHECK YOUR NEIGHBOR v o = 0 m/s (rest) t = 4 s d = ____ m a = g = 10 m/s 2 d = ½ *a*(t) 2 d = ½ *(10 m/s 2 )*(4 s) 2 d = (5 m/s 2 ) * (16 s 2 ) d = 80 m Freely Falling

39 © 2010 Pearson Education, Inc. EXAMPLE: A ball thrown straight up into the air begins to slow down (constant acceleration) until it stops in mid air at a height of 34 meters. How long does it take to reach this height? (NOTE: the same equation for free-fall is the same) d = 34 m t = _____ s a = g = 10 m/s 2 Freely Falling d = ½ *a*(t) 2 2*d = (t) 2 a t = (2*d/a) 1/2 t = (2*34 m / 10m/s 2 ) 1/2 t = (68 m / 10m/s 2 ) 1/2 t = (6.8 s 2 ) 1/2 t = 2.6 s d = 34 m Is this answer REASONABLE?

40 © 2010 Pearson Education, Inc. Free FallHow Far? at t = 1 s; d = ½ *a*(t) 2 = ½ (10 m/s 2 ) (1 s) 2 = 5 m at t = 2 s; d = ½ *a*(t) 2 = ½ (10 m/s 2 ) (2 s) 2 = 20 m d = ½ *a*(t) 2 at t = 3 s; d = ½ *a*(t) 2 = ½ (10 m/s 2 ) (3 s) 2 = 45 m at t = 2.6 s; d = ½ *a*(t) 2 = ½ (10 m/s 2 ) (2.6 s) 2 = 34 m Δd change in distance Δt change in time v = = = m/s dtdt v = = m/s Δv v f - v o change in velocity m/s Δt t change in time s a = = = = = m/s 2 d = ½ *a*(t) 2 = ½ *(m/s 2 )*(s) 2 = m a = g = 10 m/s 2 = 9.8 m/s 2 (for freely falling objects)

41 © 2010 Pearson Education, Inc. Figure 3.8 t 0 = 0 s v 0 = 30 m/s t 1 = 1 s v 1 = 20 m/s t 2 = 2 s v 2 = 10 m/s t 3 = 3 s v 3 = 0 m/s t 4 = 4 s v 4 = -10 m/s t 5 = 5 s v 5 = -20 m/s t 6 = 6 s v 6 = -30 m/s t 7 = 7 s v 7 = -40 m/s (negative = direction) a = g = 10 m/s 2 a = ΔvtΔvt v f = v o + a*t t 4 = 1 s t 5 = 2 s t 6 = 3 s t 7 = 4 s d = ½ *a*(t) 2

42 © 2010 Pearson Education, Inc.


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