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Sicn0565lfaErdanges spced from 40 mph to 64 mph Eminutes, its average acceleratonuesrour:12 milcfhouriminulc4 milesthour{minuteMlccthour minuicMlcsnourminucQuestion...

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Sicn0565lfaErdanges spced from 40 mph to 64 mph Eminutes, its average acceleratonuesrour:12 milcfhouriminulc4 milesthour{minuteMlccthour minuicMlcsnourminucQuestion0.5 ptsMEWSC Divsang minus sien: object slow down?indicate thz directionsof tlic velocity Jnd Jcelefationobject which the following sirujtions does theLICSL aIL Loltcclneealive velocilyEgalivc Jccc cralicmIrctc lnone 0I [CSCAIL Lomeciposilivevclocily MC Raenecelennncgalive vclocily ad po live ecceleralunNoir

Sicn 0565 lfaErdanges spced from 40 mph to 64 mph E minutes, its average acceleraton uesrour: 12 milcfhouriminulc 4 milesthour{minute Mlccthour minuic Mlcsnourminuc Question 0.5 pts MEWSC Divsang minus sien: object slow down? indicate thz directionsof tlic velocity Jnd Jcelefation object which the following sirujtions does the LICSL aIL Loltccl neealive velocily Egalivc Jccc cralicm Irctc lnone 0I [CSCAIL Lomeci posilivevclocily MC Raenecelenn ncgalive vclocily ad po live ecceleralun Noir



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The average speed of a vehicle in miles per hour on a stretch of Route 134 between 6 A.M. and 10 A.M. on a typical weekday is approximated by the expression $$20 t-40 \sqrt{t}+50 \quad(0 \leq t \leq 4)$$ where $t$ is measured in hours, with $t=0$ corresponding to 6 A.M. Over what interval of time is the average speed of a vehicle less than or equal to $35 \mathrm{mph} ?$

For this problem were given the average speed of a vehicle um between six and 10 a.m. So it's gonna be 20 t minus 40. Route T. Plus 50. Yes. Um T is measured in hours. So we want to know over what interval of time the average speed of the vehicle is less than or equal to 35 MPH. So we want to see when this is less than or equal to 35. By the way that we can do this is by graphing this here and then we'll look at when this is less than equal to 35. So y equals 35 miles an hour. And we see it's less than less than or equal to this um at 0.25 So that time would be 6 15 up until 2.25 hours after 10 o'clock. So it's going to be um to I'm sorry after six o'clock so it's going to be to 8 15. So between 6 15 and 8:15 a.m. And that's when the speed is less than 35 miles an hour.

Mhm. Here, we have the scenario, we see that john drives 16 kilometers west at a speed of 90 kilometers per hour shown right here. He then drives eight kilometers south at a speed of 80 kilometers per hour shown here. Finally, he drives 34 kilometers of South east at a speed of 100 kilometers per hour shown here. So if we assuming troubles at a constant velocity during each segment, we ask ourselves what is the change in velocity during the trip And to find that we're going to need to know the final condition and the initial condition. Let's try that out. At this point here, he's driving at 100 kilometers per hour. This is the final condition. At this point we have an X. Component and we have a way component since we know that he was driving directly southeast, we can consider this angle to the 45 degrees. So to find the Acts, which is the adjacent side will say that X is equal to the links, which is 100 kilometers per hour times the co sign of 45 degrees. And we find that why is equal to 100 kilometers per hour times the sine of 45 degrees. It's the opposite side, plugging that into a calculator 100 times the coastline. 45. We find the X. Component to the equal to 70 0.7 kilometers per hour. And likewise we find why to be negative 70.7 kilometers per hour. Why is it negative? Because it's in the minus Y. Direction directed downward. So now that we have the final condition will take a look at the initial conditions at this point, it's moving directly east or west. At this point, we can assume the X. To be equal to 90 kilometers per hour. Why? Because it's directed in the X. Direction. And this is going to be negative since it's in the negative X. Direction and we can assume the Y two B zero because there's no vertical movement, there's no north or south movement here, zero corners per hour. So what is the change in velocity? Well, to do that, we're gonna need to find the change in X. And the change and why? The change in acts is going to be the final, which is 70 0.7 minus the initial, which is negative 90 the change and why it's going to be the final, which is negative 70 27 minus the initial which is your So we find that the change in Y is 160.7 the change in X rather than the change in Y is negative 70 great center. Both of these are kilometers per power. So now that we found the magnitude of the change in X. And the change and why we can use those components to find a total change in velocity. So to do that, we're going to take the square root right of this change in X squared plus this change in y squared. We'll do that the square root of 160.7 squared plus 70.7 squared. We get a value 175 0.6 kilometers per hour. The change in velocity between 0.1 and point to point to is the final condition. 0.1 is the initial condition. This value is our change in velocity during the trip. And so now when we ask ourselves what was the average acceleration during this trip, we already have the change in velocity. Now we need the change in time to do that. We're gonna need to figure out how much time was spent during this trip. So we're gonna have to use each of these different segments to figure out how long it took. Sam let's start with the beginning 16 kilometers at 90 kilometers per hour. If we have a distance and velocity we can divide that distance by the velocity to get time for changing time is equal to that speed that velocity divided by or rather that distance that change in D. Which I'll call distance divided by the speed I'll call s. So for the first one we have a distance of 16 kilometers at 90 kilometers per hour. We get zero 0.1 eight hours. For the next part we have eight kilometers at 80 kilometers per hour. Eight divided by 80 gets us 0.1 hours. And for the third segment we have 34 kilometers at 100 divided or 100 kilometers per hour, which gets us 0.34 hours adding these. All together, we find that the total change in time is equal to 0.62 hours. We have our change in velocity here to find our acceleration, our average acceleration during this trip. We just have to get this change in velocity very, very that's changing time. When we do that 175.6 kilometers per hour, divided by 0.62 hours, we find an acceleration of 284 right three 0.2 rather kilometers per hour. Screen. Thank you.

So here we know that the, um for party we want to find the average speed. We know that the total distance traveled is going to be 500 kilometers and then we just simply need to find the total times The total distance traveled equals 500 kilometers. And then we can say that going there we can save the average speed or the average velocity would be equal to delta X of one divided by delta t someone we can solve For delta tease of one, this would be equal to delta X of one, of course, divided by the average velocity in the first portion of the trip. This is equaling 250 kilometres divided by 95 kilometers per hour. So this is giving us 2.632 hours for the second time. Delta T's up to this would equals Delta, except to which is again 250 kilometres on, divided by the east average velocity in the second portion. So this would be 250 kilometers divided by 55 kilometers per hour and this is giving us 4.5 for five hours Now There's a one hour lunch break so the delta T total would be equal to 4.545 Plus, what do you say for 4.545 plus 2.632 plus one So one hour lunch break and this is giving us 8.177 hours. So to find the average speed, this would be equal to Delta X Total divided by doubt the tee total and this would go 500 kilometers divided by 8.177 hours, giving us approximately 61 kilometers per hour. And then to find the for part B, we want to find the average velocity. However, the average velocity here, um, this is a round trip. So when you're when you're travelling round trip, you were you are ending up in the same place as you started. So your displacement, which defines velocity, will actually be zero. So if your displacement a zero your average velocity for the entire round trip would, of course be able to zero meters per second. That is the end of the solution. Thank you for watching

In part we need to calculate the average speed, the average speed. This is equal to distance table. We had a great time taken. So the distance civil is the difference between the meter reading. So it will be close to 124, minus 12 If I have to you worried where I am taken, The time taken is two hours. So it will be equals two 16 4 divided by two. It is equal to I did too. You learned with the but our So this is the average speed. In part B. We need to calculate, we need to calculate the velocity. The velocity is equal to displacement divided by time taken. So here since the car is starts from from from a point and returns again to the same point, so displacement is equal to zero divided by time taken. So the last will come out with zero.


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