Key Areas covered Equations of motion for objects

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

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

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

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

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

2010 Qu: 1

2005 Qu: 2

2005 Qu: 2

The Equations of Motion

The Equations of Motion

1) v = u + at 2) s = ut + ½at² 3) v²

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

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=

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

Equations of motion Travelling horizontally

Example 1 Q A car travelling at 20 ms-1 accelerates at 5 ms-2 for

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

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

2008 Qu: 21

2004 Qu: 22

2004 Qu: 22

Equations of motion Travelling vertically

Equations of motion Travelling vertically

What about acceleration due to gravity? • When an object is shot up 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

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

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)