FACULTY OF SCIENCE Physics Bridging Course Chapter 2 MOTION IN A STRAIGHT LINE School of Physics

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Motion along a straight line

Motion: ‘kinematics’ ›Position, distance and displacement x1x1 x2x2 distance = 12m displacement = 7m position 1 position 2 direction = E 20º S

Displacement ›“As the Crow Flies” ›Displacement -∆ x = x 2 – x 1 ›Vector quantity — size and direction, “four metres to the left”

Displacement. Worked example Negative distance doesn’t make much sense... but negative displacement has meaning! x 1 = 4mx 2 = –7m0 Displacement ∆x = x 2 – x 1 = (–7) – 4 = –11m (that is, 11 metres to the left)

Measuring things Before we can measure things, we need - a reference point - a scale (including definition of positive and negative)

Speed and Velocity ›Speed = (distance travelled)/(time) ›Velocity = (displacement)/(time) Vectors Can define average speed and velocity, and instantaneous speed and velocity Not vectors

Speed and Velocity. Worked example ›If it takes 2s, x1x1 x2x2 distance = 12 m speed =12 / 2 = 6 m.s -1 displacement = 7 m velocity = 7 / 2 = 3.5 m.s -1

Average vs Instantaneous ›Racing cars 2.6 km circuit ›1 lap: 1 min 18 s ›Max speed 194 km.h -1

Average Speed and Velocity. Worked example ›Average Speed s av = (total distance)/∆t › Average velocity v av = ∆x/∆t = (x 2 – x 1 )/( t 2 – t 1 ) ›Calculate s av and v av for figure 2.2 to see that they are, in general, very different!

Problem Solving If you want to solve physics problems, you need a good acronym. How about DDESC: Diagram — draw one. Specify zero and +ive direction Data — write it all down, and identify the unknown that you need to find out Equation — what’s the best equation(s) to use? Solve — Make the unknown the subject of the equation Calculate — Only use the calculator at the last step, to minimise the damage from calculation errors

Problem Solving Now CHECK to make sure your answer is not SUS: Significant Figures — right number? Units — are they correct and SI? Sensible — does the answer make sense?

Graphs (for 1-D)

t x x 1, t 1 x 2, t 2 Slope = (x 2 – x 1 )/(t 2 – t 1 ) Graph tricks ›(x 2 – x 1 )/( t 2 – t 1 ) is also the slope of the straight line joining the two points

Instantaneous speed and velocity ›The averages are over a period of time — also talk about speed or v ‘right now’, or ‘at a specific time’ ›Mathematically, we calculate the instantaneous velocity by letting ∆t → 0 v = lim ∆t→0 ∆x/∆t = dx/dt ›This is slope of the tangent to the position-time curve at that point ›Instantaneous speed = magnitude of instantaneous velocity

t x x, t Slope of tangent = dx/dt Velocity is the derivative of the position Graph Tricks

Questions?

x, t Chapter 2 continued: Motion along a straight line

Questions from last lecture/tutorial?

This lecture Complete Chapter 2 ›Acceleration ›More on graphs - integration ›Deriving equations ›Worked problems

Acceleration Acceleration is rate of change in velocity Like velocity: average or instantaneous acceleration a av = ∆v/∆t = (v 2 – v 1 )/(t 2 – t 1 ) a = lim ∆t→0 ∆v/∆t = dv/dt Since velocity is a vector, so is acceleration Acceleration is slope of velocity-time graph Acceleration is the derivative of the velocity

Graphs of Motion ›Since v is slope of x-t graph, and ›a is slope of v-t graph, We can get a lot of information from graphs of motion! See Figures 2.3, 2.4 and 2.5

Graphs of Motion t x

t x v

Graphs of Motion t x v a

If you have a position-time graph, you know how to find the velocity, acceleration from calculating the slopes but what if you knew about the velocity or acceleration of an object in motion, but not the position?

t v t1t1 t2t2 v  t = v(t 2 –t 1 ) = base x height = Area v =  x/  t (e.g. km.h -1 ),  x = v  t (km.h -1 x hours travelled) v Area = displacement in time  t

t v Each area = displacement in that time period

t v

t v

t v

In general, displacement = area under graph t v

t v ∆x = ∫ t1 t2 v dt In general, displacement = area under graph

And velocity = area under acceleration-time graph ∆v = ∫ t1 t2 a dt t a

Constant acceleration If the slope of the velocity curve is constant then acceleration is constant … a = dv/dt =  v/  t t v a

Deriving a velocity equation Let’s look at the info we already have and try to put it in terms of v a = dv/dt =  v/  t = (v – v 0 )/(t – t 0 ) Rearrange to get v 1 = v 0 + a (t – t 0 ) or v = v o + at Your first DERIVATION!

Deriving an equation for x Displacement is the area under the velocity curve, so let’s integrate: v = v o + at(1) then x = ∫v dt = ∫(v 0 + at) dt and so x – x 0 = v 0 t + ½ at 2 (2)

Deriving an equation without t Combine the two equations we already have v = v o + at (1) then x = ∫v dt = ∫(v 0 + at) dt and so x – x 0 = v 0 t + ½ at 2 (2) Substitute (1) into (2) and you get v 2 = v a(x – x 0 ) (3)

Three Equations of Motion v = u + at(1) s = u t + ½ at 2 (2) u 2 = v 2 - 2as(3) SUVAT

s = u t + ½ at 2 u 2 = v 2 - 2as v = u + at a = (v – u)/t t = (v – u)/a

Worked example. Car

Free-fall acceleration Galileo showed that All objects fall under the force of gravity with the same acceleration: a = g ≈ 9.8 ms -2

Yeah, right. As if! Drop a ball and a feather from the same height and they hit the ground at the same time?

Free-fall acceleration Galileo showed that under ideal conditions*, all objects fall under the force of gravity with the same acceleration: a = g ≈ 9.8 ms -2 * No messing about with air resistance, dropping from roughly same distance from centre of the Earth... actually, g varies slightly as you travel around

Worked exmaple. Tennis Ball

Summary of Ch 2 ›Position x and displacement  x ›Speed and velocity v ›Average velocity v av =  x/  t = slope of straight line between end points of motion ›Instantaneous velocity v = dx/dt = slope of tangent to graph of position

Summary of Ch 2 › Average acceleration a av =  v/  t › Instantaneous acceleration a = dv/dt › Areas under curves: displacement = area under velocity graph, velocity = area under acceleration graph

Summary of Ch 2 › Constant acceleration equations: - v = v 0 + at -  x = v 0 t + ½ at 2 - v 2 = v a  x › Acceleration due to gravity: g = 9.8 ms -2

Questions from last lecture/tutorial?