5

2. Normal and Tangential Motion (15 points) At a given instant the train engine at E has speed of 15 mls and acceleration of IOm/s? acting in the direction show Det...

Question

2. Normal and Tangential Motion (15 points) At a given instant the train engine at E has speed of 15 mls and acceleration of IOm/s? acting in the direction show Determine the rate of increase in the train speed and the radius of curvature of the pathFigure Problem

2. Normal and Tangential Motion (15 points) At a given instant the train engine at E has speed of 15 mls and acceleration of IOm/s? acting in the direction show Determine the rate of increase in the train speed and the radius of curvature of the path Figure Problem



Answers

At a given instant the train enginc at $E$ has a speed of $20 \mathrm{m} / \mathrm{s}$ and an acceleration of $14 \mathrm{m} / \mathrm{s}^{2}$ acting in the direction shown. Determine the rate of increase in the train's speed and the radius of curvature $\rho$ of the path.

Well after a given time lost, he has increased to V is equal to eight. Bless the cedar 80.5 multiply by 20 which is equal to 18 m per second. That means the centripetal X relation is equal to B squared, divided by role, which is equal to 18 square, divided by 400. Roy's for integrated that's ready is so eccentric. Relax. Relation is equal. Do 0.81 m per second screen here. Finally, the magnitude of facts relation is equal to root a C Square, less tangential X Relations Square, which is equal to root. It's a 2.81 square with 0.5 square, so is equal to 0.95 m per second, I swear.

Okay we have two parts of this question for part B. I've already written a formula out but we'll talk about that when we get there. So first let's start with part A. We're going to find the velocity acceleration and speed of the particle travelling along R. Of T. So to find the velocity we are going to take the derivative. So we will have a negative to sign of two T. In the I. Direction and then a. To co sign of two T. In the J. Direction and the zero in the K. Direction. Our acceleration is the next derivative. And so we will have a negative four sign of two T. In the I. Direction And a negative for sign of two T. In the J. direction. To find our speed. We're going to find the magnitude of velocity. So we can go ahead and say the square root of that negative to society of two T. Squared. So we get a four sine squared of two T. Plus a four co sine squared of two T. Well if you have a sine squared plus coastline square, that is one. So it ends up being squared. A four which is the value of two. So are A. Of T. Notice how you get A. F. T. It is the derivative of the magnitude of velocity. So because our magnitude of velocity or speed has the value of two, it's a derivative is zero. So now we can go on and we're going to find the normal component of acceleration. So for that we need to do the cross product between our prime and our double prime. Well that really is the cross product of the N. A. So when you do cross products notice that both of our jake or K components are zero. So that's going to eat zero out. Both are I and J. Component of the cross product. So we're really doing our K. Component. So when we do that we end up with that negative to sign of two T. Times the negative four sign of two T. So we get a eight sine squared of two T. Now we do minus but our second piece is negative. So for that we get a co sign of two T. So again um this is going to um you know, again you sign. Squarepants, coastline squared is one, so you get that value of eight and then we divide by the magnitude of the velocity, which is our speed. So eight divided by two, ends up with a value of four.

Hello. In this question, we have a train that is moving on a rail until it faced the horizontal curve And it starts to slow down in this curve from 90 km/h until it reached 50 km/h at the end of the curve. The time that it takes the train to reach from this velocity to this velocity was 15 seconds. And also we know that the diameter of this curve Or the radius is 150 m. So we're asked to calculate the acceleration at the point when the train was moving at 50 km/h. So first of all, we know that in this case we have to accelerations one that is radio and points towards the center of the curve and the other acceleration will be parallel to the trains, velocity or in the data direction if we use this notation. So let's first determine the acceleration that or the tangential exploration which is responsible for actually decreasing the speed rather than changing the velocity. So here we need to calculate 18 gentle and basically this is delta V by delta T. Because assuming that the exploration was constant throughout the entire 15 seconds. So here the final velocity was 50 and the initial velocity was 90. And let's convert this to be in meters per second instead of kilometers per hours. So we need to multiply by 1000 to convert to meters and divide by 3000 and 600 to convert two seconds. And this happened over 15 seconds. And Blocking on these numbers on a calculator, we will find that the answer is negative no .74 m/s. And that's in the direction of data. And of course the negative sign means that it's opposing the motion of the train. So here, if we assume that the motion of the train was is in this direction. So now we know that the exploration or the tangential acceleration factor is pointing in that way. So that's here negative open 74 m per second square and looks at a square here as well. Now let's calculate the centripetal acceleration basically by doing this squared over R and at this point the velocity was 50 kilometers per hour. Again, let me convert this 50 kilometers per hour to be in meters per second instead, And we need to square this entire value and divide by 150 m, which is the radius of the curve to find out that the answer is 1.29 m per second square, and this is always pointing toward the center of the curve. So this is this points in that direction and the valley is 1.29 m per second square to get the exploration, which is a vector. To get the value of this vector. Just using the trig Identity. That taking the square out of the squares of both explorations Plus. Open 7 4. of course the negative sign doesn't matter here because we square it anyway to find the answer which is 1.48 m per second square. So that's the answer. And this is the magnitude of the exploration as a whole vector and were asked to determine its direction as well. So from here, we know that in if we did If we take 10 data, so let's draw this here and let me assume that this one and this one this one. So that's fair to. So getting the 10 in verse of let me put the horizontal component over here, which is 0.74 and divided by 1.29. To find out that data is actually 29.8 for decrease, so that's it. The angle is 29.8 for from the radius of the curve. It depends on how on what in which in which direction Will you throw your figure? But it doesn't matter just to say that is 29 84° above the radius of the current.

We have a curve of radius. Uh huh. Mm. 200 m with there's an angular acceleration of 15 times 10 To the negative 3rd power radiance per second squared. Oh, erase angular acceleration. Okay. And an angular speed. 0.0 five radiance per second was the total acceleration of the train. Okay, well the centripetal acceleration is V squared over R. The tangential acceleration is oh, for our so the total acceleration, the magnitude. Hey now I still have the stuff down here. The magnitude is going to be oops square root of a c squared plus a t squared. So putting that in a calculator, square root uh, B squared over R squared. I better declare these variables. R. 200. Alfa equals £1.5. 10 To the negative 3rd B Equals went through five. So it's square it of the quantity V squared over R. And then I want to square that. Plus the quantity alpha are I don't want to square that. So that gives me 0.3 00 meters per second squared B. What's the angle relative to the radio direction? So let's say that it's here. This is the radio. No, no, no, that's not the radio direction. My bed, this is the radio direction. And so this would be tha tha so data would be the inverse tangent of the opposite over the high pot. No, opposite over adjacent. The opposite would be the tangential acceleration. The adjacent would be the centripetal acceleration. So putting that in the calculator. Okay. Of and gentle acceleration which is alpha are over V squared over our 1.570, that's in Radiance change out 2°. Uh huh. It's almost entirely in the tangential direction Gives Me 90°. Let's check Question # 47 Ha ha! I see where I made a mistake. I wrote this up here as V but it's supposed to be omega V is going to be omega are so um, down here, I need to write, well, I could write omega squared are here. Okay, let's do this again. So erase that and erase that. Okay, let's try this again. Um, so V is omega are, so that's just 10 m per second. Okay, that gives me .583 down here And then down here it gives me 30 1° and those are the correct answers.


Similar Solved Questions

5 answers
Stockbroker has kept daily record of the value of particular stock over the years and finds that prices of the stock form normal distribution with mean of 58.52 with standard deviation of $2.38.Refer to Exhibit 5-1- What percentage of the distribution lies below s7.42?
stockbroker has kept daily record of the value of particular stock over the years and finds that prices of the stock form normal distribution with mean of 58.52 with standard deviation of $2.38. Refer to Exhibit 5-1- What percentage of the distribution lies below s7.42?...
4 answers
What is the locus of points satisfying Iz _ 21/ 1z - 22/? What is the locus of points satisfying 21| 22 12| 9. Prove the inequality Iz1w1 + Z2w2l < VIz1l? + [227 iwl W2/
What is the locus of points satisfying Iz _ 21/ 1z - 22/? What is the locus of points satisfying 21| 22 12| 9. Prove the inequality Iz1w1 + Z2w2l < VIz1l? + [227 iwl W2/...
5 answers
Us LS andtas Excekauky t dtkrm n ( P # + Lnchm '5 Cmntinuru at Y -; Flx)-[ 2-3* 'f Y 2-1 (ie Pem6) 7*-1+3 !877
Us LS andtas Excekauky t dtkrm n ( P # + Lnchm '5 Cmntinuru at Y -; Flx)-[ 2-3* 'f Y 2-1 (ie Pem6) 7*-1+3 !877...
5 answers
LUMOChoose one: sin hydronium CCtin the alkene CCoin the alkene 0-Ho* in hydronium CCt* inthe alkene 0-Hoin hydronium
LUMO Choose one: sin hydronium CCtin the alkene CCoin the alkene 0-Ho* in hydronium CCt* inthe alkene 0-Hoin hydronium...
5 answers
[5]5. Determine whether (1,-2,2,4) is in the span of the vectors (1,0,1,0), (1,0,-2,1), (2,0,1,2}. Justify.[8] 6. Let V be the set of all ordered pairs of real numbers, with the following addition and scalar multiplication operations on V: For i = (U,uz) and V = (V,Vi) u+V = (U Uz, V +Vz) and kii = (0,kuz-Determine whether Axioms 7 and 9 hold: Justify.
[5]5. Determine whether (1,-2,2,4) is in the span of the vectors (1,0,1,0), (1,0,-2,1), (2,0,1,2}. Justify. [8] 6. Let V be the set of all ordered pairs of real numbers, with the following addition and scalar multiplication operations on V: For i = (U,uz) and V = (V,Vi) u+V = (U Uz, V +Vz) and kii =...
5 answers
In a plot of first ionization energy versus atomic number for Periods 2 and 3 , "dips" occur at the $3 mathrm{~A}$ and $6 mathrm{~A}$ elements. Account for these dips.
In a plot of first ionization energy versus atomic number for Periods 2 and 3 , "dips" occur at the $3 mathrm{~A}$ and $6 mathrm{~A}$ elements. Account for these dips....
5 answers
Solve.$$|a+2|=13$$
Solve. $$ |a+2|=13 $$...
1 answers
The owner of a bicycle repair shop forecasts revenues of 160,000 dollars a year. Variable costs will be 45,000 dollars , and rental costs for the shop are 35,000 dollars a year. Depreciation on the repair tools will be $\$ 10,000$. Prepare an income statement for the shop based on these estimates. The tax rate is 35 percent.
The owner of a bicycle repair shop forecasts revenues of 160,000 dollars a year. Variable costs will be 45,000 dollars , and rental costs for the shop are 35,000 dollars a year. Depreciation on the repair tools will be $\$ 10,000$. Prepare an income statement for the shop based on these estimates....
1 answers
Integrate each of the given functions. $$\int \frac{y d y}{4 \sqrt{25-16 y^{2}}}$$
Integrate each of the given functions. $$\int \frac{y d y}{4 \sqrt{25-16 y^{2}}}$$...
5 answers
Oono bntboaFha Proualan D2nioy Ribto Lna
Oono bntbo a Fha Proualan D2nioy Ribto Lna...
5 answers
1 46 7 H6 CHzo ChjClko'" cHz
1 46 7 H6 CHzo Chj Clko'" cHz...
5 answers
33%0 Part (b) What is the force in the y-direction Fy 33% Part (c) Neutons? What I5 / thc magnitude ofthe force Tcctor, in N?
33%0 Part (b) What is the force in the y-direction Fy 33% Part (c) Neutons? What I5 / thc magnitude ofthe force Tcctor, in N?...
5 answers
Due Thursday; Octoher 10Let R+ denote the set of positive real numbers. Prove that(-Tr) ={0} reR+
due Thursday; Octoher 10 Let R+ denote the set of positive real numbers. Prove that (-Tr) ={0} reR+...
5 answers
Ccnsider DNA Polymerase and RNA Polymerase of cukaryotes:what is the function of DNA Polymerase (no matter prok or euk)?b what Is the function of eukaryotic RNA Polymerase?what is the template for DNA Pol?what is the template for RNA Pol?describe the reaction that DNA Pol carries Out; including, does require primer?describe the reaction that RNA Pol carries out; including does require primer?
Ccnsider DNA Polymerase and RNA Polymerase of cukaryotes: what is the function of DNA Polymerase (no matter prok or euk)? b what Is the function of eukaryotic RNA Polymerase? what is the template for DNA Pol? what is the template for RNA Pol? describe the reaction that DNA Pol carries Out; including...
5 answers
Consider the following boundary value problem for the wave equation: Utt Uzr = 0 0 < x < L 0 < t<T u(z,0) = u(r,T) = 0 0 < x < L u(0,t) = "(L,t) = 0 0 < t<TAssume thatfor some integers m and n, show that family of functions defined as below are solutions to the PDE(K is arbitrary):nTr mtt I = K sin( - sin( -II) Show if solutionis irrational then there ezists no solution beside trivial zero
Consider the following boundary value problem for the wave equation: Utt Uzr = 0 0 < x < L 0 < t<T u(z,0) = u(r,T) = 0 0 < x < L u(0,t) = "(L,t) = 0 0 < t<T Assume that for some integers m and n, show that family of functions defined as below are solutions to the PDE(K i...

-- 0.021949--