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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 $t=14 \mathrm{s}$.

Physics 101 Mechanics

Chapter 11

Kinematics of Particles

Section 3

Graphical Solutions

Moment, Impulse, and Collisions

Dwight I.

October 7, 2021

in the vt curve, how did you know the graph touches t=4s? is it just an approximation?

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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'll label the area from 0 to 6 seconds. We'll call it a one, and that'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'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'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'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'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'll do this as follows. We have the initial position. It'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'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'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'll call it D is 32 last 48 last eight, which is a total distance that the particle travels off 88 ft.

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