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A uniformly accelerated car passes three equally spaced trafficsigns. The signs are separated by adistance d = 20 m. The car passesthe first sign at t = 1.5 s, thes...

Question

A uniformly accelerated car passes three equally spaced trafficsigns. The signs are separated by adistance d = 20 m. The car passesthe first sign at t = 1.5 s, thesecond sign at t = 3.3 s, and thethird sign at t = 4.9 s.(c) What is the magnitude of the acceleration of the car?

A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 20 m. The car passes the first sign at t = 1.5 s, the second sign at t = 3.3 s, and the third sign at t = 4.9 s. (c) What is the magnitude of the acceleration of the car?



Answers

A car accelerates at $2.0 \mathrm{m} / \mathrm{s}^{2}$ along a straight road. It passes two marks that are $30 \mathrm{m}$ apart at times $t=4.0 \mathrm{s}$ and $t=5.0 \mathrm{s}$ What was the car's velocity at $t=0$ s?

In this problem. We have two cars and B which are initially next to each other at the origin at rest and they begin accelerating with different accelerations and stop at different times and velocities. And we want to know what the acceleration versus time, velocity versus time and the distance versus time graphs look like so first for the acceleration versus time graph. This is most straightforward for a it's initially at 4 m, so for meters per second square and it's like that for 10 seconds and then it maintains a constant velocity. So the acceleration is then zero for be. It's at 5 m per second, squared 5 m per second squared and it does this for some amount of time, which will figure out right now. So it is accelerating at five million stuffing sweat and has a final velocity of 25 m per second. So we know that the is equal to be not plus acceleration times time, the initial velocity of zero, so we can just plug in our final velocity of 25 m per second, which is equal to our acceleration of 5 m per second times the time and we find that the time is just 25/5, which is just five seconds. So this should be a half as long as a is accelerating for so that's five seconds. And then again, it's just zero because it is maintain a cost in velocity. Next, we want to know the, uh, with the velocity grabs look like So we know what the slope of thes are going to be initially because the slopes are going to just be equal to the acceleration. And then once they're done accelerating, they're going to be a flat line since they're maintaining a constant velocity. Will do be first, since we already know what the final velocity is going to be. So for be we're gonna have some line. And this line is going to be such that the final velocity is 25 m per second and the time at which this occurs is five seconds. And then after that, it's going to be a straight line. So that's what B will look like. And for a Why don't we go ahead and find the final velocity using this equation so we know the acceleration of the time. The initial velocity of zero. So like we did last time vehicles 80 which is just four times 10 which is it's time, which is just 40 meters per second. So a it's going to have a more shallow uh, it's gonna have a less steep line. Can be because the acceleration is lower. So it's going to look something like that if we can draw a straight line and these air both should both be starting at deep oops at the origin and then or pretend this is the access. But this is going to be 40 m per second and the time which again pretends on the access, is going to be at 10 seconds. And then after that, it's going to be a straight line. No, the last for this problem is what does the position look like over time. So the what broadly is goingto happen? Is there going thio? They're accelerating initially, so they have some like parabolic shape that and then they have a constant velocity. So then they will just be a straight line so blue, since it has a steeper acceleration, is going toe look like this and then it's gonna have some constant velocity that it's gonna go off into space and red is going to have a more shallow Rapolas. Since the acceleration is less, it's going to reach a gonna be going for longer, and then it's going to have a higher speed than so it's going to be a straight line after that. That's going to be steeper than blue, since the final velocity is greater. And if we say this is around where it stops or I guess maybe this is around where it stops decent accelerating. So this is 10 seconds where it stops accelerating. And for Blue, where it stops accelerating is five seconds, and we want to know. In addition to this, what the distance between the cars is when T is equal to 15 so T equals 15 so that's somewhere over here. So what we can do for this is we can use the equation that the change in the distance is going to be the initial velocity times time plus one half of the acceleration Times T squared for an Excel constant acceleration. And then after that, the, uh, change in distance. It's just going to be the velocity times the time. So we can add those up, then Thio get the total distance changed for both of them. So we'll start with a so delta us. The initial velocity is going to be zero is going to be one half times four times 10 squared, plus a final velocity of 40. And this is going to be for until t equals 15. So 15 minus 10 is five, two plus five times 40. So Delta s for car A is one half times force. That's too times 100 plus five times 40 is 20 times 10 is 200. So this is two times 200 which is just 400 m. At T equals 15 and for Blue Delta s and one half five times the time it's accelerating for which is five seconds squared. Plus, that's final velocity of 25 m per second time to the time that it's going. That velocity which is 10 seconds so that will give us five cubed is 125 So 125 over to plus 250 which we can simplify this as 1 25/2 is 50 plus 25 or two which is 12.5 plus 2 50. So this is going to be 300 plus 12.5, which is just 300 in 12.5. So that's our answer for adults A s and finding the difference between them is just 400 minus 312.5, which is going to be 112.5 was gone. 10 there nine. That's not a nine Sophie and eight. It's more like being paid, but we need a 0.5, so the distance between the two is 87.5 meters, and that's the solution.

The day Engine chillax relation is equal to do be divided five et attending to lax Relation is equal to rate of change of velocity. Right. And you re is equal to DS, divided by city. When is different chatted with respect to time Then we get velocity And from here we can right DT is equal to d s d minded by re yeah, since 80 is equal to on TV divided by D TV before 80 is equal to g B divided by DT and DTs Dia's divided by TV DS divided by B All right on this implies age Tangential X relation is equal to I'll be times you d be divided by D. S Andi, we have ah TV divided by D s Legner I don't here DVD divided by D s is equal to minus sooner 0.15 It's minus 0.15 on tending to lax. Relation is equal to 25 25 minus 10.15 No 0.1 five into minus zero point 15 15 Right. So well, we have a s is equal to all 51.5 meter, 51.5 meter. So we have miss here as well. So s is able to 50 51.5 meter and therefore are tending to lax Relation is equal to minus 2.59 minus two point 59 meter per second square. And when s is equal to when s is equal to 51.5 meter and musical too. 17.275 meters per 2nd 17.2 75 meter per second. Well, we have ah rule. Right. Well, uh, Rome is equal to one less divi divided by the X deep my divided by D X clear into three, divided by two Divided by Andi Square Why divided by D X trail and de by divided by D X eyes equal to minus do divided by 625 times X x is equal to 50 on and this implies state D by divided by D. X is equal to minus 0.16 right and d square d squared. Why divided by are the X Square is able to minus 0.0 32 and then rule is equal to 324 0.576 meter right sequel to 300 UH, 24 point 576 meter and being with eight V is equal to real times. Peter Daut and pleaded not, is equal to be divided by role. So Peter Don't is equal to 17 point 275 divided by 324 point 576576 So a pita don t is equal to zero corn. Siegel concede a 00.0 0.532 five three to Radian per second, radiant for second and normal X relation. Normal X relation is a cold, too. Are all times Peter, not square well, which is equal do 0.919 me, too, per second square. Now Total X relation is equal to root. Dandan two X Relations Square Place normal Next Relations Fear, which is equal to 2.75 music or second square

Okay, We have one car moving at a constant speed of 33 meters per second. Another car starts from rest. So be not, is zero and accelerates at some acceleration. And they the two cars meet at a later point at 2.5 kilometers or 2500 meters from their initial position. So we need to calculate that acceleration. Um, it might be helpful to visualize this in a graph. One car is moving at a constant speed up. This is a position versus time graph one car, the first car car, one moving at a constant speed. The other car starts from rest and speeds up. So here's where they meet at the 2500 meter position after having been at the same place at the same time at the beginning of the problem. So we don't have time, but it is easy to find the time. If you look at Car one car one. The distance distance for car one is just velocity times T because there's no acceleration. So 2500 meters is equal to 33 meters per second, claims the time rearranging that we have a time interval of 75.76 seconds. So now we have a tea. So our next job is to calculate what the acceleration must be for car to in order to reach a 25 meet 100 meter distance. So let's use this equation. Me not tea plus 1/2 a T squared. So plugging my numbers in for car to 2500 meters, the initial is zero, so I don't even have to put that into the equation. 1/2 a time. 75.76 seconds squared so I can do the ultra 75.76 square divided by two. And then I guess I could multiply both sides by two and then divide both sides by 75.76 seconds squared and we have an acceleration of 0.8712 meters per second squared rounded to two significant figures 0.87 meters per second squared

So at the end of the acceleration period, the velocity, the final would be going the initial velocity which we know to be zero plus the acceleration times t during the acceleration or the time of acceleration. And so this would be equaling 1.5 meters per second squared multiplied by 5.0 seconds and this giving us 7.5 meters per second. Now the this would be the initial velocity for the breaking period. So for part A, we can say after breaking, we can say velocity final equals velocity initial plus the acceleration times the time of the break and so this would be giving us 7.5 meters per second, plus the acceleration of negative 2.0 meters per second squared multiplied by 3.0 seconds, and we have a final velocity equaling 1.5 meters per second. This would be our final answer for part A for part B, Then the total distance traveled. Ah would be equaling essentially the average velocity times T during the acceleration period, plus the average velocity times T during the break in period, so this would be equaling 7.5 meters per second, divided by two most supplied by 5.0 seconds. And this would be plus 1.5 meters per second plus 7.5 meters per second, all divided by two times 3.0 seconds. And we find that then Delta X total is equaling 32 meters. This would be our final answer for part B. That is the end of the solution. Thank you for watching.


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