Speaking the Same Language in Physics and Math

Marci K. Harvey

West Forsyth High SchoolClemmons, NC

Thanks To the Following:

• North Carolina Center for the Advancement of Teaching (NCCAT)

• Heather King, Math teacher• Daphne Marshall, Math teacher• Kurt Telford, Principal, West Forsyth

Today’s Objectives

• Demonstrate how students can make a connection between math and physics

• Compare solutions to typical problems with math and physics formulas

• Justify the benefit to students of teaching both methods

The Problem…

• Students do not make connections between physics concepts and math concepts, even if they take the courses concurrently!

• x = Vcost• y = Vsin()t-gt2/2

Speaking the Same Language…

Physics Unit: Freefall motion, kinematic equations

Math Unit: Parametric equations, quadratic equations

Problem: A ball is dropped from 100m on Earth, the moon, and Jupiter. How long does it take the ball to reach the surface at each location?

Forms of the Equations

• Math (parametric) Equations for Motion:y = ax2 + bx + c

• Physics Equations for Motion:yf = 1/2at2 + vit + yi

Parametric Solution…

gearth = -9.8 m/s2

gmoon = -1.6 m/s2

gjupiter = -26 m/s2

To set up calculator:• Turn “stat plot” OFF• Mode Par, Dot, Simul

• Y= X1T = 1 (moves object away from y-axis)

• Enter Y1T equation

– h(t) = -4.9t2 + Vot + ho

• Set “window”– Tmin = 0 s– Tmax = you decide– Tstep = 0.3 s recommended

100m

Parametric Solution…

gearth = -9.8 m/s2

gmoon = -1.6 m/s2

gjupiter = -26 m/s2

To set up calculator:

• Y= X2T = 2 for moon and X3T = 3 for Jupiter

• Enter Y2T and Y3T equations

• Graph and watch!• “Trace” to find time when

ball hits surface

100m

Physics Solution…

gearth = -9.8 m/s2

gmoon = -1.6 m/s2

gjupiter = -26 m/s2

Kinematic solution:

yf = yi + vit + ½gt2

0m = 100m + ½(-9.8m/s2)(t2)

t = 4.5 s for Earth

t = 11.2 s for moon

t = 2.8 s for Jupiter

100m

How about motion in two dimensions?

Physics Unit: 2D motion

Math Unit: Parametric Equations/Projectiles

Problem: An outfielder throws a softball 28 m/s at an angle of 55o from the ground. How long will the ball be in the air? Will it make it to the catcher, 80 m away?

Forms of the Equations

• Parametric Equations for Motion:x(t) = vtcos y(t) = vtsin – 1/2gt2 + h

• Physics Equations for Motion:

xf = xi + vt

yf = 1/2at2 + vit + yi

Parametric Solution…Parametric solution:

• Calculator settings– Degree mode

• Enter equations– X1T = 28T(cos55)

– Y1T = 28T(sin55) – ½(9.8)T2

• Set window:– Tmin = 0; Tmax = 5; Tstep = 0.1– Xmin = 0; Xmax = 100– Ymin = 0; Ymax = 50

55o

28 m

/s

Parametric Solution…Parametric solution:

• Graph and watch!• “Trace” to find time when ball

hits surface

Answer:

t = about 4.6 s

x = about 74 m

55o

28 m

/s

Physics Solution…Kinematic solution:

Vf = Vi + at

0 = 23 m/s + (-9.8m/s2)(t)

t = 2.3 s to the top of the parabola

t = 4.6 s for entire flight

x = vt = (16m/s)(4.6s)

= 73.6 m

It does not reach home.

55o

28 m

/s

Vx = (28 m/s)(cos55) = 16 m/s

Vy = (28 m/s)(sin55) = 23 m/s

Other Areas to Consider…

• Vector addition in physics

• Dot product and Law of Cosines in math

• x,v,a graphs/relationships in physics

• Integrals/derivatives in math

Questions? Final thoughts?

• Your ticket out the door!

• My email: mharvey@wsfcs.k12.nc.us

• Phone: 336-712-4400