a-particle-moves-in-a-straight-line-with-a-constant-acceleration-of-4-ftls-2-for-6-s-zero-accelerati

Question

A particle moves in a straight line with a constant acceleration of $-4$ ftls $^{2}$ for 6 s, zero acceleration for the next 4 s, and a constant acceleration of $+4 \mathrm{ft} / \mathrm{s}^{2}$ for the next $4 \mathrm{s}$. Knowing that the particle starts from the origin and that its velocity is $-8 \mathrm{ft} / \mathrm{s}$ during the zero acceleration time interval, (a) construct the $v-t$ and $x-t$ curves for $0 \leq t \leq 14 \mathrm{s},(b)$ determine the position and the velocity of the particle and the total distance traveled when $t=14 \mathrm{s}$.

Keshav Singh · Accepted Answer

for this problem on the topic of cosmetics of particles were given the emotion off a particle during a 14 2nd journey. It has constant acceleration for its first part of the journey, zero acceleration for the next part and again, constant acceleration. For its final part, we know the time for each part of the journey as well as its velocity. During the zero acceleration phase, we are asked to construct the V, T and X curves for the total journey, as well as the position and velocity of the particle and its total distance traveled after he total time has elapsed so well. First, construct the acceleration time curve and we&#x27;ll label the area from 0 to 6 seconds. We&#x27;ll call it a one, and that&#x27;s minus 24 ft per second and the area under the curve from 10 to 14 seconds. The final constant acceleration phase, which is 16 ft a second. So labeling it a one and a two. We have a one two B minus 24 ft second, and the area under the exploration time graph for the second constant acceleration phase is 16 feet a second, and you can see that the units often accept off a velocity. So let&#x27;s look at part A. And first we need to construct the VT cough. Now we know that at six seconds the six is given to a minus eight feet a second, and so we can calculate the initial velocity. V, not a T is equal to zero. This is simply V six minus a one, the area under the 80 graph for the first part of the journey. This is minus eight minus minus 24 which gives an initial speed off 16 ft a second. Now, at 10 seconds, we have eaten to be again minus 80 ft per second since this previous constant from 6 to 10 seconds. And the final velocity V 14 is equal to whatever the speedy&#x27;s at time 10 plus the area under the acceleration time curve from 10 to 14 seconds. Remember, the area on an acceleration time curve gives a D velocity, so that&#x27;s minus eight feet per second. Class 16, which gives the speed at 14 seconds to be 8 ft the second. And so from here we can sketch the VT curve as follows and from the Bt curve, we label the areas. A three, a four, a five, a six and a seven, as we have in the diagram. So these areas the areas under DVT cough give us position. That&#x27;s can calculate these areas from the curve. So we firstly have a three, which is the area of a triangle to be 32 feet. The area a full two B, minus 8 ft a five is minus 32 ft. A six is minus 8 ft and a seven final area is equal to 8 ft. So now we have the areas and will use this to calculate or to construct the exchequer brother. Okay, now to construct our X T curve. Yeah, we&#x27;ll do this as follows. We have the initial position. It&#x27;s not to be zero. He positioned after t is equal to four seconds. X four is equal to x note. Lastly, area a three. And so this is simply 32 feed. Since the initial position of zero he positioned at teasing of six seconds at six is equal to X four plus area a fort and this is 24 ft. X 10 is equal to x six plus eight five. It&#x27;s and this is minus 8 ft after 12 seconds. The position of the particle X 12 is equal to x 10 Okay plus area in six under the meeting draft, which gives us position x 12 to B minus 16 ft and finally the final position x 14 at T 0. 14 seconds is x 12. So last area A seven, which is minus eight feet. Now, with this information, we can plot the curve as follows. And so from hour X T curve has shown we can calculate the total distance traveled. Okay, so the total distance traveled. We&#x27;ll do this in sections from 024 seconds. We have distance traveled. The one is equal to the absolute value off 32 minus zero, which is a total distance off 32 feet between times four seconds and 12 seconds. We have a distance D two, which is the absolute value off minus 16 minus 32 ft, which gives us a total distance traveled or 48 feet, and between a time of 12 seconds and 14 seconds. We have a distance. The three, which is the absolute value off minus eight minus minus 16, which gives us a final distance in the last two seconds off 8 ft and therefore the total distance traveled. We&#x27;ll call it D is 32 last 48 last eight, which is a total distance that the particle travels off 88 ft.

Zulfiqar Ali · Answer

Well, position is a function of time. Could be regiones integral 40 times. DT So, uh, this into girl is equal to three d squared minus 60 times DT. Since we equals Treaty Square minus 60 and solving this integral, our position is a function of time Musical two D, Q Minus free T square. Let&#x27;s see where sees a constant well, it Time t is equal to zero. Uh, position is zero Q minus three to CDO square place scene and therefore si is equal to four. So C is equal to four, right? Since ah c equals all right position in Time T is equal to zero, So seize equal to four and further, like music rollup um, we have position is a function of time is equal to a cube minus three t square less four feet and position when time physical four seconds is equal to forgive Minus three into four square this fuller musical to 20 peach right, So we have that&#x27;s not is equal to four feet and s one is equal to 20 feet and some office not And this one is equal to 24 feet. No, let&#x27;s find out acts relation we don&#x27;t which is, um TV divided Fine TT eyes equal to x relation of the function of time which is equal to 60 minus six feet or second squared. So x relation when time physical two seconds His six into two mine six Matiz ical to six beat were second square.

Keshav Singh · Answer

for this problem on the topic of cosmetics of particles were told that a particle is moving in a straight line with acceleration has shown in our acceleration time chart. We know that the particle started the Origen with a given initial speed on. We are asked to construct the V T and X curves for the total emotion between zero and 18 seconds, as well as the position and velocity of the particle as well as the total distance after a time of 18 seconds has elapsed. So if we plot the 80 graph and we label the area a one which is between times zero and eight seconds, eight to the area under 80 graph for 8 to 12 seconds and a three the area under the 80 curve from 12 to 18 seconds we confined these areas and we know the areas under acceleration time graph give give us philosophies. So a one is equal to minus 0.75 times eight, which is minus 6 m per second. Uh huh. Area A to this is to multiplied by four, which is 8 m per second. In the final area, a three is equal to six times six is a rectangular areas and that gives us 36 m per second. Now we can find the values off the velocity at the indicated times as follows. So we not the initial velocity V no is minus 2 m per second. The velocity at eight seconds B eight is equal to V. Not last area a one and we know a one, so this is minus 8 m per second. So that&#x27;s minus two plus minus six, which is minus eight. The speed of particle at 12 seconds, V 12 is able to V eight plus area a to which is zero. So the speed of the particle for 12 seconds zero the final velocity of the particle. The 18 is equal to V 12, plus the area A three, which is 36 m per second. And now we can use this information to first sketch the VT curve. Using straight line portions over the constant acceleration periods on the graph looks as follows and from our Bt graph as we have shown, we can label areas a four, a five and a six for each part of the journey. Now we confined these indicated areas as follows. So the first area A four is equal to half into minus two minus eight times eight, which is minus 40 m for the area under a meaty graph, gives us position A five is equal to a half times minus eight times for which gives us a displacement off minus 16 m and finally the area the final part of the motion A six. It is ah half into 36 times six, which gives us 100 and eight meters. So now we can find the values of the position at the times indicated. We know the initial position. X not is equal to zero the position after eight seconds using the Vita graph his ex note plus the area in four which is minus 40 meters. The position after 12 seconds X 12 is equal to x eight less the displacement, which is a five and this is minus 56 m, Really. At the end of the journey x 18, the final position the particle is able to x 12 last area A six, which is positive 52 m now, using these values weaken sketch the next curve as follows now with our position time graph sketched as follows. We confined our position and our velocity at the Times indicated so. The position off the particle AT T is able to 18 seconds is X 18 and X 18 we have shown is 52 meters. So that&#x27;s the position of the particle at the end of its journey. The velocity at this time we got from our VT curve A T is equal to 18 seconds. We labeled a V 18 and this is 36 m the second and finally the total distance traveled by the particle Hungary. We&#x27;re called D indeed is 56 plus 100 and 8 m, so the distance in each direction, which gives us a total distance of 100 and 64 meters so it calculated, or the quantities that we require.

R M · Answer

a problem. 23 ah be average equal to here. Immigration from one before or three. You you lost Bt genes equal through their times. Ah, the integration for that is equal to 3/4 people are four minus What you By substitution from one at Fuller it will be equal to 789 Over or so, the final result would be loosely to 63 over over four Which m b the the average? It&#x27;s equal Ah to ah one over E minus A. Or this will be minus IHS which is cheaper to third voice of institution 100 minus seven. G 31 as we even read is equal to one and being equipped with.

Karuna Ramroop · Answer

all right. So given this function of the velocity of this particle, if we want to find the total displacement, we can think of that as the position function we&#x27;re being asked to find The total displacement between T equals zero in T equals Chu. So the idea is, if the particle was going this fast with respect to time, how far did it travel? Um, between zero seconds and two seconds. So we know that position is the integral of velocity. So to find the total displacement, we&#x27;re going to do the integral of our velocity Oceans of four minus two tea with respect to t. And we&#x27;re not just going to find the function, we&#x27;re going to find the displacement from zero to choose. So this is how we would set up our, um, internal expression. And now just all this. Um, we&#x27;re just getting used powerful. We have to separate terms so the integral of four is gonna be 40. The integral of duty is going to be chiu ti to you. The power there&#x27;s an invisible one is our exponents. We&#x27;re going to you. I had one and then we&#x27;re going to divide by this quantity we get in the exploded and then this is going to be evaluated on zero, 22 So then our next step is too simple by our terms. So before t, we&#x27;re gonna leave alone because there&#x27;s nothing much to do there and then or the exponents This is t raised you the one plus one which is just t raised to the power to And this is too divided by one post horn, which is just you and two divided by two is just gonna be one. So this is 40 minus t squared, evaluated from zero Chu. So did you bring this to the next page 40 Minus t square? Why you waited from 0 to 2, and now we&#x27;re going to plug in those limits. So we&#x27;re going to begin or chew first. So four terms too minus two squared, and then we&#x27;re going to subtract the expression that we get when we plug in zero. All right, so four times to you is gonna be eight minus and then to swear is gonna before. And then this entire second expression disappears because this is your Oh, and this is zero and the difference of zero. And zero was just zero. So our entire expression just becomes eight minus four, which is just for so. That means that our total displacement is for

Problem

A particle moves in a straight line with a consta…

Problem 61 Medium Difficulty

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Ferdinand P. Beer, E. Russell Johnston, Jr.,David F. Mazurek, Phillip J. Cornwell, Brian P. Self

Vector Mechanics For Engineers Statics and Dynamics

Elyse G.

Andy C.

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Marshall S.

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Vector Mechanics For Engineers Statics and Dynamics

Elyse G.

Andy C.

Liev B.

Marshall S.

Math Review - Intro

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Ferdinand P. Beer, E. Russell Johnston, Jr.,David F. Mazurek, Phillip J. Cornwell, Brian P. SelfVector Mechanics For Engineers Statics and Dynamics

Elyse G.

Andy C.

Liev B.

Marshall S.

Ferdinand P. Beer, E. Russell Johnston, Jr.,David F. Mazurek, Phillip J. Cornwell, Brian P. Self

Vector Mechanics For Engineers Statics and Dynamics