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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88

Problem 37 Medium Difficulty

A car is traveling at a constant speed of 33 $\mathrm{m} / \mathrm{s}$ on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two
cars meet for the first time at the next exit, which is 2.5 $\mathrm{km}$ away?

Answer

$a=.87 \frac{m}{s^{2}}$

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Physics 101 Mechanics

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Chapter 2

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Video Transcript

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

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John D. Cutnell, Kenneth W. Johnson

Physics

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