Key Areas covered

1. Key Areas covered • Equations of motion for objects moving with constant acceleration in a straight line

2. What we will do today: • Revise the definition of acceleration. • State the equations of motion. • Carry out calculations on equations of motion.

3. Acceleration

4. Acceleration Acceleration = change in velocity time taken = final velocity – initial velocity time taken a = v – u t a: acceleration (ms-2) v: final velocity (ms-1) u: initial velocity (ms-1) t : time taken (s)

5. What does this mean? ACCELERATION IS THE CHANGE IN VELOCITY PER UNIT TIME. An acceleration of 1 ms-2 means the velocity of the body changes by 1 ms-1 every second. Units are metres per second per second or ms-2.

6. 2010 Qu: 1

7. 2005 Qu: 2

8. The Equations of Motion

9. 1) v = u + at 2) s = ut + ½at² 3) v² = u² + 2as u – initial velocity at time t = 0 v – final velocity at time t a – acceleration of object t – time to accelerate from u to v s – displacement of object in time t

10. These equations only apply to uniform acceleration in a straight line. The vector quantities displacement, velocity and acceleration have direction associated with them, and so they will have a positive or negative sign depending on their direction.

11. Method for Tackling Problems Write down all the symbols like this: u = v = s = a = t = Fill in all numbers & values given in question. If there are two directions, use this diagram for + and – values: Choose the best equation to suit the problem. Up + Left - Right + Down -

12. Equations of motion Travelling horizontally

13. Example 1 Q A car travelling at 20 ms-1 accelerates at 5 ms-2 for 2 s. How far does the car travel during the 2 s? A u = 20 ms-1; a = 5 ms-2; t = 2 s; s = ? Use s = ut + ½at² s = 40 + 10 s = 50 m

14. Example 2 Q A train travelling at 45 ms-1 decelerates to 15 ms-1 at 2 ms-2. How far does the train travel while it is decelerating? A u = 45 ms-1; v = 15 ms-1; a = -2 ms-2; s = ? (note –ve sign) Use v² = u² + 2as 15² = 45² - 4s 4s = 45² - 15² s = 450 m

15. 2008 Qu:21

16. 2004 Qu:22

17. Equations of motion Travelling vertically

18. What about acceleration due to gravity? • When an object is shot up vertically in one dimension (ie no horizontal travel) acceleration due to gravity has to be considered. • On Earth, a = 9.8 ms-2, and this always acts downwards. • Therefore, if an object is launched vertically up, we have two directions (we have both +ve and –ve). • Questions don’t usually contain this value, you are expected to know it!

19. Example 1 • A ball is launched vertically upwards at a velocity of 15 ms-1. What is the height of the ball after 2s? • (note accn has not been mentioned here, however you are expected to know it is involved). • Choose directions: ↑ = +ve and ↓ = -ve • u = 15ms-1 t = 2s a = - 9.8ms-2 s = ? • s = ut + ½at2 = (15 x 2) + (½ x (-9.8)x22) = 30 + (-19.6) s = 10.4 m

20. 2001, Qu: 3

21. Experiment • Find your reaction time using a ruler • Everything is moving downwards so take ↓ = +ve (if you use –ve you will still get the same answer – the important bit is you are consistent). • a = 9.8 ms-2 • s = distance travelled by ruler (in m) • u = 0 • s = ut + ½ at2 • s = ½ at2 (as u = 0)

22. Questions • Activity sheets: • Equations on Motion (suvat) • Page 6 – 8 • Answer all questions in class jotter