Chapter 4: Accelera.on

Transcription

1 Chapter 4: Accelera.on

2 Accelera.on the rate at which velocity changes (increases or decreases). changing velocity = accelera6on constant velocity = zero accelera6on

3 a = v t = v f - v i t a = average accelera6on (m/s 2 ) v = change in velocity (m/s) t = 6me taken for change (s) v f = final velocity v i = ini6al velocity

4 Example: A car accelerates from 20 km/h to 80 km/h in 10 seconds. Calculate the average accelera6on. (Assume 2 sig. figs. for all measurements) v f = 80 km/h = 80 km/h x 1000 m x 1h = 22 m/s 1 km 3600 s v i = 20 km/h x 1000 m x 1h = 5.6 m/s 1 km 3600 s t = 10 s a =? a = v f v i t = 22 m/s 5.6 m/s 10s = 16.4 m/s 10s = 1.64 m/s 2 = 1.6 m/s 2

5 Example 1: A runner accelerates from 0.52 m/s to 0.78 m/s in s. What is her accelera6on? v f = 0.78 m/s v i = 0.52 m/s t = s a =? a = v f v i t = 0.78 m/s 0.52 m/s s = 0.26 m/s s = 5.2 m/s 2

6 Example 2: A car accelerates from rest at a rate of 50.0 cm/s 2 for 12.5 s. How fast is it now moving? v f =? cm/s v i = 0 cm/s t = 12.5 s a = 50.0 cm/s 2 a = v f v i t v f = v i + at = 0 cm/s + (50.0 cm/s 2 )(12.5s) = 625 cm/s

7 Example 3: A turtle wants to accelerate from 2 mm/s to 8 mm/s. How long does it take, if its maximum accelera6on is 3 mm/s 2? v f = 8 mm/s v i = 2 mm/s t =? a = 3 mm/s 2 a = v f v i or t t = v f v i a = 8 mm/s 2 mm/s = 6 s 3 mm/s 2

8 Accelera.on and Slope Accelera6on is equal to the slope of a velocity- 6me graph v t

9 v A B A: constant accelera6on B: greater constant accelera6on t

10 Posi.ve Accelera.on Constant accelera6on produces a straight line (increase of velocity is the same for each unit of 6me). Changing velocity produces a curved line (increase in velocity is not the same for each unit). v v t t

11 Nega.ve Accelera.on Slowing down produces nega6ve accelera6on Straight line when decrease in velocity is the same for each unit of 6me. Curved line when decrease in velocity changes with 6me. v v t t

12 Zero Accelera.on Velocity remains constant. v t

13 Change in Direc.on If direc6on changes, the velocity changes and therefore, there is accelera6on. i.e., Ferris wheel rotates at a constant speed in a circular mo6on, but the direc6on changes, resul6ng in accelera6on.

14 Changing Accelera.on Curve on a V- T graph v t Average accelera6on = slope of the straight line joining two points on a curve of a V- T graph

15 Changing Accelera.on v t Instantaneous accelera6on = slope of the tangent of a point on the graph.

16 1. At what 6mes is the accelera6on zero? 9.2s, 12.8s and 18.0s

17 2. What is the accelera6on for the first 7.0 s? (5.2-0 m/s)/7.0 s = 0.74 m/s2

18 3. What is the average accelera6on for each of the following 6me intervals? a) 5.0 to 15.0 s? b) 9.0 to 13.0 s? c) 15.0 to 20.0 s? ( m/s)/10.0s =0.39 m/s2 ( m/s)/4.0s = m/s2 ( m/s)/5.0s =0.76 m/s2

19 4. What is the accelera6on for each of the following 6mes? a) 15.0 s? b) 11.0 s? c) 17.0 s? ( m/s)/( s) =2.3 m/s2 ( m/s)/( s) = m/s2

20 Velocity with Constant Accelera.on Velocity- 6me graph is a straight line Slope of the line = accelera6on a = v f v i t v f = v i + at v f = final velocity (m/s) v i = ini6al velocity (m/s) a = accelera6on (m/s 2 ) t = 6me (s)

21 Ex#1. A car with a velocity of 5.0 m/s, accelerates at a rate of 5.0 m/s 2. What is the velocity of the car at a 6me of 3.5 s? v i = 5.0 m/s v f =? a = 5.0 m/s 2 t = 3.5 s v f = v i + at = 5.0 m/s + (5.0 m/s 2 )(3.5s) = ( ) m/s = 22.5 m/s = 23 m/s

22 Ex#2. A supersonic jet that is flying at 145 m/s is accelerated uniformly at the rate of 23.1 m/s 2 for 20.0s. What is its final velocity? v i = 145 m/s v f =? a = 23.1 m/s 2 t = 20.0 s v f = v i + at = 145 m/s + (23.1 m/s 2 )(20.0s) = ( ) m/s = 607 m/s

23 Displacement During Constant Accelera.on

24 Displacement: Given Velocity and Time displacement is equal to the total area under the line of a velocity- 6me graph. During constant velocity from rest: v ave = d t d = v ave t v f d = ½ (v f + v i )t v i t

25 Ex#1. How far does a dragster travel in 6.00 s, accelera6ng steadily from zero to 90.0 m/s? d =? v f = 90.0 m/s v i = 0.00 m/s t = 6.00 s d = ½ (v f + v i )t = ½ ( m/s)(6.00s) = ½ (90.0 m/s)(6.00s) = 270 m = 2.70 x 10 1 m

26 Ex#2. Two skateboarders accelerate steadily from 4.5 m/s to 11.5 m/s in 6.0s. How far do they travel? d =? v f = 11.5 m/s v i = 4.5 m/s t = 6.0 s d = ½ (v f + v i )t = ½ ( m/s)(6.0s) = ½ (16.0m/s)(6.00s) = 48 m

27 Displacement: Given Accelera.on and Time If v f = v i + at and d = ½(v f + v i )t then d = ½(v i + at + v i )t d = ½(2 v i + at)t d = ½(2 v i t + at 2 ) d = v i t + ½ at 2

28 Ex#1. A skier accelerates at 1.20 m/s 2 down an icy slope star6ng from rest. How far does she get in 5.0 s? d =? v i = 0.0 m/s a = 1.20 m/s 2 t = 5.0s d = v i t + ½ at 2 = (0.0m/s)(5.0s) + ½(1.20 m/s 2 )(5.0) 2 s 2 = 0 m m = 15m

29 Ex#2. What is the accelera6on of an object that accelerates steadily from rest, traveling a distance of 150 m over 10.0 s? a =? d = 150 m v i = 0.0 m/s t = 10.0s d = v i t + ½ at 2 a = 2(d- v i t) t 2 a = 2[150m (0.0)(10.0)m] (10.0s) 2 a = 300 m 100 s 2 a = 3.0 m/s 2

30 Ex#3. How long does it take an airplane accelera6ng from rest at 5.0 m/s 2 to travel 300 m? a = 5.0 m/s 2 d = 300 m v i = 0.0 m/s t =? s d = v i t + ½ at 2 ; since v i =0 m/s d = ½ at 2 t = s or 11 s

31 If Displacement: Given Velocity and v f = v i + at Accelera.on and t = v f - v i a

32 Ex. A bullet accelerates at 6.8x10 4 m/s 2 from rest as it travels the 0.80 m of the rifle barrel. a) How long was the bullet it the barrel? a = 6.8x10 4 m/s 2 v i = 0.0 m/s d = 0.80 m t =? d = v i t + ½at 2 d = ½at 2 = s = 4.9x10-3 s

33 Ex. A bullet accelerates at 6.8x10 4 m/s 2 from rest as it travels the 0.80 m of the rifle barrel. b) What velocity does the bullet have as it leaves the barrel? v f =? v i = 0.0 m/s a = 6.8x10 4 m/s 2 d = 0.80 m v 2 f = v 2 i + 2ad = 0.0 m/s + 2(6.8x10 4 m/s 2 )(0.80m) v 2 f = m 2 /s 2 v f = m/s = 3.3x10 2 m/s

34 Accelera.on due to Gravity Galileo showed that all objects fall to earth with a constant accelera6on, if air resistance can be ignored. g is the symbol for accelera6on due to gravity. On the surface of the earth, g = 9.80 m/s 2 (varies slightly, depending on distance from centre of earth) Assuming no air resistance, all accelera6on formulas apply to fall objects (subs6tute g for a)

35 Example: The Hellevator ride at Playland falls freely for a 6me of 1.8 s. a) What is the velocity at the end of the 6me? v i = 0 m/s v f =? t = 1.8 s g = 9.8 m/s 2 v f = v i + at = 0.0 m/s + (9.8m/s 2 )(1.8s) = m/s = 18 m/s

36 Example: The Hellevator ride at Playland falls freely for a 6me of 1.8 s. b) How far did it fall? v i = 0 m/s t = 1.8 s g = 9.8 m/s 2 d =? d = v i t + ½gt 2 = (0.5)(9.8m/s 2 )(1.8s) 2 = m = 16 m