Information about MOTION- Velocity, Acceleration,graphs

MOTION- Velocity, Acceleration, graphs

2. Learning objectives Students will be able to • Define speed • Define velocity • Calculate speed and velocity

3. WARM UP

4. Speed • Distance moved per second or rate of change of distance • It is scalar quantity • Unit-m/s • Average speed=Distance travelled/time taken

5. Velocity • Velocity measures rate of change of displacement • Average velocity=Displacement/time taken • It is a vector quantity. • Unit- m/s

6. CONTINOUS ASSESSMENT 1.What does the speedometer of a car measure? 2.A car travels a distance of 2oom and returns to the original position after 5min.What is the speed and velocity of the car? 3.What is the eqn. for speed? 4.What is the eqn. for time taken?

7. Final assessment 1.a)Define speed b) Define velocity (page42)

8. HOME WORK • Page.33 Example.2

9. GRAPHS Displacement –time graph a) Body at rest t S t(s) 0 1 2 3 4 S(m) 2 2 2 2 2

10. b)Body moving with steady velocity t s t(s) 0 1 2 3 4 S(m) 0 5 10 15 20 Displacement Gradient= Time Gradient of displacement –time graph gives velocity. Steeper the line ,larger the velocity.

11. c)Accelerating body t 0 1s 2s 3s 4s s 0m 1m 4m 9m 16 a=2m/s² t s Gradient at different time instants are different, as the velocity is changing at every instant. Gradient at any instant gives the instantaneous velocity

12. d)Decelerating body t s

13. Velocity –time graph a) Body moving with steady velocity t v

14. b)Body moving with steady acceleration t v t(s) 0 1 2 3 4 v(m/s) 0 5 10 15 20 acceleration Changeinvelocity • Gradient= Time • Gradient of velocity –time graph gives acceleration. • Steeper the line ,larger the acceleration.

15. c)Body moving with steady deceleration t v t(s) 0 1 2 3 4 v(m/s) 20 15 10 5 0 Changeinvelocity • Gradient= • Gradient of velocity –time graph gives acceleration. • Steeper the line ,larger the deceleration. acceleration deceleration Time /

16. Area under the velocity –Time graph gives the displacement Calculate the total displacement of the given body using the given velocity time graph.

17. OA-accelerated motion; Displacement = 100m AB-Steady velocity ; Displacement = 400m BC—decelerated motion; Displacement = 50m Total displacement = 100+400+50=550m

18. Page43. Qn . 8

19. a) acceleration= gradient of OA =change on y/change on x =y2-y1/X2-X1 =2-0/3-0 =2/3 =1.5m/s² b) Decelerates, its velocity becomes zero and it reaches the maximum height. c)At point C the velocity of the lift is zero. The height=Displacement=area under OABC = ½ × 3 × 2 + 6 × 2 + ½ × 3 × 2 =3+12+3=18m. d)The lift starts moving down. First it accelerates in the downward direction.Then it starts decelerating and come to rest e) 18-6=12m

20. PAGE 106

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