Unit 03“Horizontal Motion”

Problem Solving Skills Assessment

Step 1: List Variables and Assign Values – using units and/or key words assign numbers to symbols

Identify which variable is being represented in a noncontextualized problem

25 m/s2____ 25 min ___ 25miles ___ 25s ____ 300 km ____ 20 hr _____ 20 mi/hr ___ 20 yr ___300 m/s ___ 5.6 km/yr ____ 15 m/s2 ____ 25 m/min East ____ 25 m ___

V

V ΔxVV

Δta

a Δt ΔxΔx

ΔtΔt

Identify variables in a contextualized problem:1. Zahida kicks a volleyball from rest giving it an acceleration of

1.5m/s2 for 14s, how fast does it go in the end? Vi=0 a= 1.5m/s2 ∆t= 14s Vf = ?

2. What is the displacement of a shopping cart that Ms Bucci pushes to accelerate at 0.5m/s2 for 3 seconds if it starts with a velocity of 4m/s?

a= 0.5m/s2 ∆x= ? ∆t= 3s Vi = 4m/s

3. Eddy rides his bicycle with a velocity of 12m/s, if he accelerates at for 4s to ends with a velocity of 38m/s, how far did it go?

Vi=12 m/s ∆t= 4s Vf= 38 m/s ∆x= ?

4. Suppose a car traveling initially at 20m/s slams on the brakes. If the brakes are applied for 5seconds and the skid marks on the ground show it took 10m to stop what rate did the car accelerate

at? Δx =10m a= ? Δt= 5s Vi= 20m/s

Step 2: “Choose Equation” Choose appropriate equation based on variable list

1. ∆t, a, Vf, Vi

2. ∆x, Vi, ∆t, a

3. Vi, Vf, ∆t, ∆x

4. ∆x, a, ∆t, Vi

 ∆x = ½(Vi+Vf)∆t

∆x = Vi∆t + ½a(∆t)2

∆x = Vi∆t + ½a(∆t)2

Vf = Vi + a∆t

Step 3: “Plug In” Plug values for variables into the equation

1. Vi=0

a= 1.5m/s2

∆t= 14sVf = ?

2.a= 0.5m/s2

∆x= ?∆t= 3sVi = 4m/s

3. Vi=12 m/s∆ t= 4sVf= 38 m/s∆x= ?

4.Δx =10ma= ?Δt= 5sVi= 20m/s

Vf = Vi + a∆t

Vf = (0m/s) + (1.5m/s2)(14s)

∆x = Vi∆t + ½a(∆t)2

∆x = (4m/s)(3s) + ½(0.5m/s2)(3s)2

 ∆x = ½(Vi+Vf)∆t

 ∆x = (12m/s+38m/s)(4s)

∆x = Vi∆t + ½a(∆t)2

10m= (20m/s)(5s) + ½a(5s)2

Step 4: “Solve” for equation above

solve the problem for the missing variable.

Step 5: “Box Answer and Check Units”

box final answer, check for correct units throughout problem

1. Vi=0

a= 1.5m/s2

∆t= 14sVf = ?

Vf = Vi + a∆t

Vf = (0m/s) + (1.5m/s2)(14s)

Vf = (0m/s) + (1.5m/s2)(14s)

Vf = 0m/s + 21m/s

Vf = 21m/s

2.a= 0.5m/s2

∆x= ?∆t= 3sVi = 4m/s

∆x = Vi∆t + ½a(∆t)2

∆x = (4m/s)(3s) + ½(0.5m/s2)(3s)2

∆x = (4m/s)(3s) + ½(0.5m/s2)9s2

∆x = 12m + ½(4.5m)

∆x = 12m + 2.25m

∆x = 14.25m

3. Vi=12 m/s∆ t= 4sVf= 38 m/s∆x= ?

 ∆x = ½(Vi+Vf)∆t

 ∆x = ½ (12m/s+38m/s)(4s)

 ∆x = ½ (50m/s)(4s)

 ∆x = 100m

4.Δx =10ma= ?Δt= 5sVi= 20m/s

∆x = Vi∆t + ½a(∆t)2

10m= (20m/s)(5s) + ½a(5s)2

10m= (20m/s)(5s) + ½a(25s2)10m= (100m) + ½a(25s2)10m= 100m + a(12.5s2)

- 100 -100m

-90m= a(12.5s2)12.5s2 12.5s2

-7.2m/s2= a

Using your group number: Check off the “step” you have shown proficiently with as you progress though the Problem Solving Skills Assessment