Equations of Motion.

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Presentation transcript:

Equations of Motion

Vectors – size (magnitude) and direction Add head to tail for relative velocity problems (eg boat on river) Subtract by adding the opposite for change problems (final – initial!!!!) Components – vertical and horizontal (projectiles)

Our basic formulae v = Δd Δt a = Δv Used for uniform motion.

Graphs of Motion

The most useful graph is The velocity time graph because from this we can get the velocity (read graph) acceleration (the gradient) distance travelled (the area under the graph)

To solve this problem A dragster travelling at a constant speed accelerates uniformly from 56ms-1 for 8s to reach a new speed of 85ms-1. What distance did it do and what was the acceleration?

Equations of motion Using a general velocity- time graph we can look at any situation with constant (or zero) acceleration and get some useful equations

Using the gradient (acceleration)

Using the area under the graph

Using tricky maths we get a 3rd equation By substituting for t from vf = vi + at into d = vit + 1/2at2 and rearranging, it can be shown that These equations are called the kinematic equations of motion

Solving problems using the kinematic eqations of motion An object is decelerating at 0.24ms-2 from an initial velocity of 22ms-1. Find the velocity of the object after half a minute

Steps to solve problem List what you know (making sure units are SI units) Decide what you want to find out Select the best equation Rearrange, substitute and solve a = -0.24ms-2 (- because deceleration) vi = 22ms-1 t = 0.5min = 30s vf = ? vf = vi + at = 22 – 0.24 x 30 = 14.8 ms-1

Try one for your self A car accelerates from rest at a constant rate of 8ms-2. How fast is it going when it is 100m from its starting position? vi = 0 ms-1 a = 8 ms-2 d = 100m vf = ? vf2 = vi2 + 2ad = 0 + 2 x 8 x 100 = 1600 vf = 40ms-1

A mathematical note In physics, when multiplying and dividing your answer should be no more accurate than the least accuarate factor (rounding to sig figs). 3601 (4sf) x 0.12567 (5sf) = 452.5 (4sf) 3601 (4sf) x 0.13 (2sf) = 470 (2sf) 100 could be 1, 2 or 3 sf. Unless told otherwise, assume it to be the most accurate (ie 3 sf)