Physics 101, Week 2 notes
Physics 101, Week 2 notes PH2213
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This 6 page Class Notes was uploaded by Deborah Oluwadamilola Eseyin on Friday August 26, 2016. The Class Notes belongs to PH2213 at Mississippi State University taught by Dr. Chun Su in Fall 2016. Since its upload, it has received 6 views. For similar materials see Physics I for Scientists and Engineers in Physics at Mississippi State University.
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Date Created: 08/26/16
Notes for Week 2 Physics 1 Instantaneous Velocity: Examples: 1. A jet engine moves along an experimental track (which we call the x axis). We will treat the engine as if it were a particle. Its position as a function of time is given by the 2 2 equation x = At + B, where A = 2.10m/s and B = 2.80m. (a) Determine the average velocity during this time interval. (b) Determine the displacement of the engine during the time interval from t1= 3.00s to t2= 5.00s. (c) Determine the magnitude of the instantaneous velocity at t = 5.00s. Solution: (a) In physics, the first thing you do when you have a question is see what the question is asking you to solve. In this question we are asked to find three things. (a) displacement (b)average velocity (c) Magnitude of the instantaneous velocity. First of all, I start off by finding the displacement since we have the value for time, A and B, so that makes it easier to find your displacement (x). T1=3s so, x 1 At +1B X 1 2.10(3) + 2.80 X 1 (2.10 x 9) + 2.80 X = 18.9 + 2.80 1 X 1 21.7m T 2 5s so, x 2 At +2B X 2 2.10(5) +2.80 X 2 (2.10 x 25) + 2.80 X 2 52.5 + 2.80 X = 55.3m 2 So therefore, displacement = x 2 x 1 D = 55.3m – 21.7m D = 33.6m (b) Magnitude of the average velocity can be calculated using the formula υ = Δx = x 2 x =155.3 – 21.7 = 33.6 Δt = t 2 t 1 5 - 3 = 2 So therefore, 16.8m/s (c) Determine the magnitude of the instantaneous velocity. First of all to find the instantaneous velocity you need to find the derivative of the x equation (x = At + B) ’ So therefore, x = 2At So all you have to do is substitute for A and t. For t we are going to use 5.00 secs since T1= t 2 5 secs. So, x = 2(2.10)(5) X = 21.0m/s 2. You are designing an airport for small airplanes. One kind of airplane that might use the airfield must reach a speed before takeoff of at least 27.8m/s, and can accelerate at 2 2.00m/s . Find the following. (a) If the runway is 150 m long, can this airplane reach the required speed for takeoff? (b) If not, what minimum length must the runway have? Solution: In this question, we were given a lot of values so in order to make things easier and way more organized. I advise that you make a table of what you have and what you don’t have. Known Wanted X 0 0 (initial distance is 0 since the plane v is at rest) V 0 0 (initial velocity is 0 since the plane is also at rest) X = 150m A = 2.00m/s 2 In this chapter, we have 3 equations called Kinetic Equations that we use to find either initial velocity, final velocity, distance, acceleration or time. V = v 0 at ……………………. (i) 2 X = v 0 + ½(at ) ……………………. (ii) 2 2 V = v +02ax ……………………. (iii) Where; v = final velocity (m/s )2 V 0 initial velocity (m/s ) X = distance (m) A = acceleration (m/s) T = time(s) Looking at all those 3 equations we need to find the final velocity and we have x, a, and v so0 we are going to use the 3 equation. 2 2 V = v +02ax 2 2 V = (0) + 2(2)(150) V = 600 Next you take the square root of both sides V = √600 V = 24.5m/s In the question we were given a supposed final velocity of 27.8m/s, but since the velocity we calculated is less than the supposed final velocity that means that the runway length is not sufficient. So that means we are going to calculate b (b) Length? rd Since we are finding the length (x) we are still going to use the 3 equation since we have all the values we need. We are going to use the supposed velocity since we are finding the required length, and also we are going to solve for x so that means the new equation 3 we are using for this problem is X = v – v 02 2a X = (27.8) – 0 2(2) X = 772.84 4 X = 193m is the required length 3. How long does it take a car to cross a 30.0m wide intersection after the light turns green, if the car accelerates from rest at a constant 2.00m/s ? Solution: First of all let’s draw out our table Known Wanted X 0 0 V? V 0 0 T? X = 30m A = 2.00 First of all I am going find V, so therefore equation 2 sounds perfect. 2 2 V = v 0 2ax 2 V = 0 + 2(2)(30) V = 120 Then take the square root of both sides. V = √120 V= 10.95m/s Then to find time we use equation 1 V = v0+ at 10.95 = 0 + 2(t) 10.95 = 2t T = 5.48secs 4. Suppose that a ball is dropped from a tower 70.0m high. How far will it have fallen after a time 1 = 1.00s,2t = 2.00s, and3t = 3.00s. Ignore air resistance. Solution: Once again we draw our table. Known Wanted T1= 1sec Y1 T2= 2 secs Y2 T3= 3secs Y3 Y = 70.0m A = +9.80 (since the y is positive, then a is positive) V 0 0 For this problem we are going to use the second equation, and change x to y since the ball is going down the y axis. So we are going to use equation 2 for all 3 times to find all 3 distances. Y 1 v t0 1½(at2) Y 1 0 + 0.5(9.80)(1)(1) Y 1 4.90m Y 2 v 0 2 ½(at2) Y = 0 + 0.5(9.80)(2)(2) 2 Y 2 19.6m Y 3 v t0 3½(at2) Y 3 0 + 0.5(9.80)(3)(3) Y 3 44.1m 5. (i) A person throws a ball upward into the air with an initial velocity of 15.0 m/s. Calculate how high it goes, and (b) how long the ball is in the air before it comes back to the hand. Ignore air resistance Solution: First of all our table: Known Wanted V 0 15m/s Y = ? A = -9.8 m/s (acceleration due to gravity T = ? is downward) V = 0 So this question has very little numbers, so we need to think about the question and figure out what equations we need to use. First of all for (a) we need to consider the time interval from the ball leaving the throwers hand until the ball reaches the highest point. So therefore, when t = 0 we have y = 0, v =015.0m/s, a = -9.80 because the ball comes to rest a little before it drops back down. Since we have this conclusion, we are going to use the third equation. 2 2 V = v +02ay (since it’s a horizontal fall) 0 = (15)(15) + 2(-9.80)y 0 = 225 – 19.6y -225 = -19.6y Y = 11.47m (b). Since the ball leaves the thrower’s hand and returns back to it so the total distance back at the thrower’s hand is zero y = v 0 + ½(at ) 0 = (15)t + 0.5(-9.80t ) 0 = 15t – 4.90t 2 0 = t(15 – 4.90t) So t = 0, 15 – 4.90t = 0 15 = 4.9t T = 3.06s and 0secs
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