Displacement with Uniform Acceleration 9:32am September 20, 2010

Scenario You’re driving on i-45 and you need to pass someone. You know you’re going 65mph and you know you only have a ¼ mile of space to pass the slowpoke in front of you. How much acceleration do you need to give your car? How much time will it take you? Will the police man up ahead notice you doing it? This is why we need to study motion formulas.

The main formula that links these all together is x = vit + ½ at2 Key Points 9:34am There are 5 main variables we can use when we discuss motion x = displacement vi = initial velocity vf = final velocity t = time a = acceleration The main formula that links these all together is x = vit + ½ at2 In this formula there are only 4 variables. This means, we will be given 3 of the variables and asked to find the 4th.

Problem Solving Strategy For each problem I’m going to write out the variables in the equation x , vi , t, a I’m going to assign the numbers I know and figure out what I’m solving for. Next, I’m going to plug in my number and then rearrange the equation to solve for the missing variable

Ex 1- 9:40am A ball rolls down an inclined plane with a uniform acceleration of 3.00 m/s2. If the ball has an initial velocity of 8.0 m/s at one point, what distance will it have traveled 9.0 seconds later? x= vi= t= a= x= vi= 8.0m/s t= 9.0s a= 3.0m/s2 x = vit + ½ at2 x = (8.0m/s)(9.0s) + ½ (3.0m/s2)(9.0s)2 x = 193.5m

Ex 2- 9:45am A student accelerates away from school towards the bus ramp at 4.44m/s2. If the student starts from rest and it takes the student 3s to get to the bus ramp, how far did the student travel to reach the bus? x= vi= t= a= x= vi= 0.0m/s t= 3.0s a= 4.44m/s2 x = vit + ½ at2 x = (0.0m/s)(3.0s) + ½ (4.44m/s2)(3.0s)2 x = 19.98m or about 20m

GP1- 9:50am A car initially traveling at 7.0 m/s accelerates at 49.0 m/s2 for 14.0 seconds. What distance does it travel during the 14 seconds? x= vi= t= a= x= vi= 7.0m/s t= 14.0s a= 49.0m/s2 x = vit + ½ at2 x = (7.0m/s)(14.0s) + ½ (49.0m/s2)(14.0s)2 x = 4900m

GP2- 9:55am An automobile with an initial speed of 11.0m/s accelerates uniformly at the rate of 7.0m/s2. Find the final displacement after 4.0s. x= vi= t= a= x= vi= 11.0m/s t= 4.0s a= 7.0m/s2 x = vit + ½ at2 x = (11.0m/s)(4.0s) + ½ (7.0m/s2)(4.0s)2 x = 100m

Independent Practice- 10:00am This is called INDEPENDENT practice for a reason. You have 4 problems to complete. Once you are finished, raise your hand and I can check it in-class. This NEEDS to be finished by the end of class. You NEED to talk NO louder than a WHISPER!

Closing- 10:25am What is the formula we can use to determine an object’s motion? What are the 4 variables in the equation and what do they stand for?