Presentation is loading. Please wait.

Presentation is loading. Please wait.

Unit 03 “Horizontal Motion” Problem Solving Hard Level.

Similar presentations


Presentation on theme: "Unit 03 “Horizontal Motion” Problem Solving Hard Level."— Presentation transcript:

1 Unit 03 “Horizontal Motion” Problem Solving Hard Level

2 Problem Solving Steps 1 st List Variables & Assign Values 2 nd Choose Equation 3 rd Plug In 4 th Solve (Simply and/or Rearrange) 5 th Box Answer

3 Hard (-) Problem Solving is about comparing your answer to a set variable. For example – if the problem says “The bus is 40m away” and asks “Can a ball you throw reach the bus?” you would: Solve for the distance you can throw the ball Compare it to how far the bus is away (40m). Hard (-) Problem Solving

4 Hard (-) You kick a soccer ball to give it an initial velocity of 11m/s. As it rolls toward the goal, friction accelerates it at a rate of -0.6m/s 2 until it stops. If the goal is 70m away, will the ball reach the goal? V i =11 m/s V f = 0 m/s a = -0.6m/s 2 ∆x= ? V f 2 = V i 2 + 2a∆x (0m/s) 2 = (11m/s) 2 + 2(-0.6m/s 2 )∆x 0 = 121m 2 /s 2 + (-1.2m/s 2 )∆x -121m 2 /s 2 = -1.2m/s 2 ∆x -121m 2 /s 2 = -121m 2 /s m/s 2 = -1.2m/s m = ∆x It will take 75.6m to stop. But the goal is only 70m away. Your friend will make the goal! Compare the goal’s distance of 70m to how far the ball can actually go.

5 Hard (+) Problem Solving is 2-steps. Includes equations from Unit 01 The two steps include F=ma V f = V i + a∆t ∆x = V i ∆t + ½a(∆t) 2 V f 2 = V i 2 + 2a∆x Hard (+) Problem Solving Use Force Equation Use one of Unit 01 Equations

6 Hard (+) 1. A 30kg dog starts from rest at the top of a muddy incline 2.0m long and slides down to the bottom in 1.5s. What net force acts on the otter along the incline?

7 1. A 30kg dog starts from rest at the top of a muddy incline 2.0m long and slides down to the bottom in 1.5s. What net force acts on the otter along the incline? F = ? m = 30kg a = ________ Need to find the acceleration 1 st !!!! Δ x = 2.0m V i = 0m/s Δ t = 1.5s a =? Δ x = V i Δ t + ½a Δ t 2 2.0m = (0m/s)(1.5s) + ½ (a)(1.5s) 2 2.0m = 0 + (1.13s 2 )a 1.77m/s 2 = a F = ma F = (30kg)(1.77m/s 2 ) F = 53.1N 1.77m/s 2

8 Hard (+) 2. How far does a 40kg carriage move if it is pushed starting from rest with a force of 260N for 3s?

9 2. How far does a 30kg carriage move if it is pushed starting from rest with a force of 260N for 3s? Δ x = ? V i = 0m/s Δ t = 3s a =________ Need to find the acceleration 1 st !!!! F = 260N m = 30kg a = ? F = ma 260N = (30kg)(a) 8.67m/s 2 = a 8.67m/s 2 Δ x = V i Δ t + ½a Δ t 2 Δ x = (0m/s)(3s) + ½ (8.67m/s 2 )(3s) 2 Δ x = m Δ x = 39.0m


Download ppt "Unit 03 “Horizontal Motion” Problem Solving Hard Level."

Similar presentations


Ads by Google