1. Ball thrown upwards and caught at same height on way down 0 A B C D Displacement Time 0 A B C D Velocity Time Upwards is positive, Initial displacement = 0 Gradient = Velocity, so: At A and C → maximum values At 0, B and D → Gradient = 0 Line ABC is straight, showing a constant negative acceleration (due to gravity) Area OAB = Area BCD as it is caught at the same height Displacement-Time Graph Velocity-Time Graph

2. v t Ball thrown in the air and caught again Ball accelerates as you throw it Ball decelerates under gravity Ball reaches the top of its motion Ball accelerates downwards under gravity Decelerates in your hand as you raise it back to d = 0 Ball hits your hand

3. Train travelling west from A to D, stopping at B and C. Assume West is positive, initial displacement = 0 and Initial velocity = 0 A B C D Displacement Time 0 Gradient of each section is the same assuming the train uses the same cruising speed Rounded curves as it starts and stops reflecting a real situation Displacement- Time Graph Velocity Time 0 A B C D Velocity – Time Graph Slight curves to lines reflecting a real situation Area under each part of curve is same if stations are equidistant from each other

4. Displacement – Time Graph A, C, E, G are all at the same displacement (floor level) Time interval gets smaller with each bounce Quite a good ‘ picture ’ of the action. What is the difference if you choose the initial displacement to be its maximum value and the floor to be zero? Ball falling to floor, bouncing 3 times. Assume upwards is positive, initial displacement = 0 All negative gradients are parallel (constant acceleration due to gravity) Areas of pairs of triangles between A’ and C C’ and E E’ and G, must be equal, (why?) What will happen if you add all the subsequent bounces? displacement A B C D E F G 0 Velocity –Time Graph 0 velocity time A A’ B C C’ D E E’ F G

5. Motion along a straight line with constant (uniform) acceleration Acceleration, a = constant displacement, s time, t Initial velocity, u Final velocity, v

6. Motion along a straight line with constant (uniform) acceleration Velocity Time u v v-u t Change in velocity is the final velocity minus the initial velocity so: Equation 1 Acceleration = change in velocity time taken = gradient

7. Motion along a straight line with constant (uniform) acceleration Velocity Time u v t Area under graph = displacement Displacement = area of trapezium = average height x base = average velocity x time = (u + v) t 2 Equation 2

8. Motion along a straight line with constant (uniform) acceleration Velocity u v t v-u Area under graph = displacement Equation 3

9. Motion along a straight line with constant (uniform) acceleration By using equations 1 and 2 we can derive Equation 4

10. Summary – write this bit down Motion along a straight line with constant (uniform) acceleration The equations of motion for: s = displacement (m) u = initial velocity (ms -1 ) v = final velocity (ms -1 ) a = acceleration (ms -2 ) t = time (s)

11. Toy Cars The aim of the game is to find things out about the motion of the toy car using the equations of motion. We can think of the motion in two parts Part 1 Part 2 We can measure: t1t1 t2t2 s1s1 s2s2

12. Part 1 t1t1 s1s1 s 1 = measured u 1 = 0 (car starts at rest) v 1 = ? a 1 = ? t 1 = measured Find a 1 and then use to calculate v 1 Part 2 t2t2 s2s2 s 2 = measured u 2 = v 1 v 2 = 0 (car finishes at rest) a 2 = ? t 2 = measured Calculate a 2

13. Part 1 t1t1 s1s1 s 1 = measured u 1 = 0 (car starts at rest) v 1 = ? a 1 = ? t 1 = measured Find a 1 using Find v 1 using or find v 1 using or

14. Part 2 t2t2 s2s2 s 2 = measured u 2 = v 1 v 2 = 0 (car finishes at rest) a 2 = ? t 2 = measured Find a 2 using

15. 1) An object travelling at 15ms -1 accelerates at 3 ms -2 for 5 seconds. Calculate (a) its final velocity (b) its displacement Equations of Motion Questions 2) A car travels at 25ms -1 for 7 seconds then slows to a halt in 5 seconds. Calculate (a) distance travelled in first 7 seconds (b) its acceleration during the last 5 seconds (c) the total distance travelled. 3) An arrow accelerates from rest at 300ms -2 through a distance of 0.5m. it then flies at steady speed 20m to a target. Calculate the total time from rest until hitting the target.

16. 1) An object travelling at 15ms -1 accelerates at 3 ms -2 for 5 seconds. Calculate (a) its final velocity (b) its displacement s=m u=ms -1 v= a=ms -2 t=s ? ? 15 3 5 a) b) 30 112.5

17. 2) A car travels at 25ms -1 for 7 seconds then slows to a halt in 5 seconds. Calculate (a) distance travelled in first 7 seconds (b) its acceleration during the last 5 seconds (c) the total distance travelled. Motion is in two parts: velocity/ms -1 time/s 25 7 12 a) 1st part: b) 2nd part:

18. 2) A car travels at 25ms -1 for 7 seconds then slows to a halt in 5 seconds. Calculate (a) distance travelled in first 7 seconds (b) its acceleration during the last 5 seconds (c) the total distance travelled. s=m u= 25ms -1 v= 0ms -1 a= -5ms -2 t= 5s s part one = 175m c) 2 nd part Add distances from 1 st and 2 nd parts and total distance = 237.5m Part 2

19. 3) An arrow accelerates from rest at 300ms -2 through a distance of 0.5m. it then flies at steady speed 20m to a target. Calculate the total time from rest until hitting the target. s= 0.5m u= 0ms -1 v= ?ms -1 a= 300ms -2 t= ?s Motion is in two parts, the first part the arrow is accelerating and the second part the arrow is at a constant speed. Find the time for the first part

20. s= 0.5m u= 0ms -1 v= ?ms -1 a= 300ms -2 t= 0.058s Find the final velocity to find the value for constant speed for the second part 3) An arrow accelerates from rest at 300ms -2 through a distance of 0.5m. it then flies at steady speed 20m to a target. Calculate the total time from rest until hitting the target. Motion is in two parts, the first part the arrow is accelerating and the second part the arrow is at a constant speed. So at constant speed: speed = distance/time so time = distance/speed = 20/17.3 =1.156s Add the times from the 1 st and 2 nd part: 0.058 + 1.156 = 1.21s