Speaking the Same Language in Physics and Math

Speaking the Same Language in Physics and Math Marci K. Harvey West Forsyth High School Clemmons, 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 = Vcos(q)t • y = Vsin(q)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) = vtcosq y(t) = vtsinq – 1/2gt2 + h • Physics Equations for Motion: xf = xi + vt yf = 1/2at2 + vit + yi

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 Parametric Solution… 28 m/s 55o

Parametric solution: Graph and watch! “Trace” to find time when ball hits surface Answer: t = about 4.6 s x = about 74 m Parametric Solution… 28 m/s 55o

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. Physics Solution… 28 m/s 55o 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

Speaking the Same Language in Physics and Math