Equations of Motion Higher Unit 1 – Section 1

Learning Intentions To be able to carry out calculations using the kinematic relationships: v = u + at, s = ut + ½at2 v2 = u2 + 2as for objects moving with a constant acceleration in a straight line.

Deriving the Equations of Motion  

Equations of motion Tips The equations of motion can only be used with objects that: Are travelling in straight line (this includes vertical) motion Have a UNIFORM (constant) acceleration. E.g. 4 ms-2 for 10 seconds, NOT the acceleration increases from 2 ms-2 to 4ms-2 in 10 seconds. We are dealing with VECTOR quantities, so we use s for displacement and v for velocity – always! A direction may be needed.

Equations Of Motion  

Some Rules……… s – displacement (m) u – initial velocity (m/s) v – final velocity (m/s) a – acceleration (ms-2 ) t – time taken (s) Always lay out what information you have been given - SUVAT! Be aware that some values are not given directly in the question

… and some more E.O.M. are often a continuation of another question style / topic. When dealing with problems that are objects moving vertically, we will need to use the acceleration due to gravity. Acceleration due to gravity ALWAYS acts down. So a will always be - 9.8 ms-2 If object is accelerating / decelerating and the question refers to Speeds / times / distances think E.O.M.!!!!!

…and finally It is important to note that anything thrown or travelling UP is positive. Anything thrown or travelling DOWN is NEGATIVE. This is for ANY vector, but more commonly displacement, velocities and accelerations.

Example of Negative vectors A person that jumps from the top of a cliff into the sea will have a negative displacement because he is going DOWN and is finishing up lower than where they started. They will also have a NEGATIVE velocity and acceleration due to gravity. Acceleration Velocity Displacement All – ve! s

Worked Example 1 A car is initially travelling at 24 ms-1. It decelerates at 4 ms-2 to a stop. How far does the car travel ? s = ? v2 = u2 + 2as u = 24 ms-1 02 = 242 + 2(-4)s v = 0 ms-1 a = -4 ms-2 0 = 576 - 8s t = s = 72 m forward is positive

Worked Example 2 A train accelerates away from a station at a constant rate. It takes 44 s to travel 340 m. What is the acceleration of the train? s = 340 m s = ut + ½at2 u = 0 ms-1 340 = 0x44 + ½a442 v = a = ? 340 = 0 + 968a t = 44 s a = 0.351 ms-2 forward is positive

Worked Example 3 A tennis ball is hit vertically upwards with a velocity of 36 ms-1. If acceleration due to gravity is taken to be 9.8 ms-2 how long does it take for the ball to reach the ground again? s = ut + ½at2 s = 0 m 0 = 36t + ½(9.8)t2 u = 36 ms-1 0 = 36t - 4.9t2 v = 0 = t(36 – 4.9t) a = -9.8 ms-2 t = 0 s or 7.35 s t = ? take off return to ground upwards is positive

Success Criteria I can successfully use the equations of motions to calculate unknown quantities for objects travelling in a straight line.