10282021 Equations of Motion 1 Equations of Motion

10/28/2021 Equations of Motion 1 - Equations of Motion Summary Notes page 1.

What are we learning today? �How use the equations of motion. v = u + at s = ut + ½ at 2 v 2 = u 2 + 2 as

What are the equations of motion? �These are the relationships that we use to describe the motion of an object numerically. �We can find them from a velocity-time graph. �Your teacher will take you through the proofs to find these relationships. N. B. You probably don’t have to be able to repeat these proofs in an exam.

An object accelerates uniformly from an initial velocity u to a final velocity v, in a time t. Velocity (ms-1) Its total displacement is given by the area under the graph. v s = ut + ½ x t x (v – u) u s = ut + ½ x t x at t s = ut + ½ at 2 Time (s)

s = ut + ½ at 2 v = u + at v 2 = (u + at)2 v 2 = u 2 + 2 aut + a 2 t 2 v 2 = u 2 + 2 a(ut + ½ at 2) v 2 = u 2 + 2 as v = u + at s = ut + ½ at 2 v 2 = u 2 + 2 as

Example 1 �A cyclist accelerates at 0. 5 ms-2 for 20 seconds to reach a final velocity of 12 ms-1. What is the initial velocity? v = u + at 12 = u + 0. 5 × 20 u = 12 – 10 = 2 ms-1

Example 2 �An object travelling at 10 ms-1 decelerates at 3 ms-2 for 3 s. Calculate its displacement. s = ut + ½ at 2 s = 10 × 3 + ½ × (-3) × 32

Example 3 �A car accelerates from rest for 5 seconds, and has a displacement of 50 m. What is its acceleration? s = ut + ½ at 2 50 = 0 × 5 + ½ × a × 52 100 = 25 a a = 100/25 = 4 ms-2

Example 4 �A car accelerates at 2 ms-2 from an initial velocity of 4 ms-1, and has a displacement of 12 m. Calculate the velocity of the car. v 2 = u 2 + 2 as v 2 = 42 + 2 × 12 v 2 = 16 + 48 v = 8 ms-1

Example 5 �A motorbike accelerates from rest at 3 ms-2 reaching a final velocity of 12 ms-1. Calculate the displacement of the motorbike. v 2 = u 2 + 2 as 122 = 02 + 2 × 3 × s 144 = 6 s s = 24 m

Structured Problems page 6, Qs 1 -16.