Key Areas covered Equations of motion for objects

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

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

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)

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 or ms-2.

2010 Qu: 1

2005 Qu: 2

The Equations of Motion

1) v = u + at 2) s = ut + ½at² 3) v² = u² + 2 as 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

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.

Method for Tackling Problems 1. Write down all the symbols like this: u= v= s= a= t= 2. Fill in all numbers & values given in question. 3. If there are two directions, use diagram for + and – values: Up + Left - this Right + Down - 4. Choose the best equation to suit the problem.

Equations of motion Travelling horizontally

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

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 + 2 as 15² = 45² - 4 s 4 s = 45² - 15² s = 450 m

2008 Qu: 21

2004 Qu: 22

Equations of motion Travelling vertically

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).

Example 1 • A ball is launched vertically upwards at a velocity of 15 ms-1. What is the height of the ball after 4 s? • (note accn has not been mentioned here, however you are expected to know it is involved). • Choose directions: ↑ = +ve and ↓ = -ve • u = 15 ms-1 t = 4 s a = - 9. 8 ms-2 s = ? • s = ut + ½at 2 = (15 x 4) + (½ x (-9. 8)x 42) = 60 + (-78. 4) s = -18. 4 m

Experiment • Find your reaction time using a ruler • Everything is moving downwards so take ↓ = -ve • a = 9. 8 ms-2 • s = distance travelled by ruler (in m) • u=0 • s = ut + ½ at 2 • S = ½ at 2 (as u = 0)