### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Physics 101, Week 2 notes PH2213

MSU

### View Full Document

## About this Document

## 6

## 0

## Popular in Physics I for Scientists and Engineers

## Popular in Physics

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.

## Similar to PH2213 at MSU

## Popular in Physics

## Reviews for Physics 101, Week 2 notes

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

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

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.