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A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.0 $\mathrm{m} / \mathrm{s}$ . The car is a distance $d$ away. The bear is 26 $\mathrm{m}$ behind the tourist and running at 6.0 $\mathrm{m} / \mathrm{s}$ . The tourist reaches the car safely. What is the maximum possible value for $d ?$

Maximum value of $d$ is 52 $\mathrm{m}$

Physics 101 Mechanics

Chapter 2

Kinematics in One Dimension

Motion Along a Straight Line

University of Michigan - Ann Arbor

Simon Fraser University

University of Winnipeg

McMaster University

Lectures

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Okay, This problem. We have a bear chasing a man and he's trying to get to his car. Uh, or a tourist could be a woman, but it's more letters. So a bear is chasing a man and he's trying to get to the car, and we're trying to find the maximum distance when they reach the car at the same time. The car has a magical portal, and obviously the man is able to get into the car lickety split in no time at all. So that's the assumptions that were making. So here it says, D max, But I'm gonna write a little bit more. I want to say D Max when, uh, time to reach the car is equal the same. That's the key. That's the insight that you need. In order to solve this problem from here, we're gonna go ahead insistently, right? The solution. The solution is going to involve the fact that we're gonna have here. The velocity is equal to the distance divided by the time, which is great. Uh, but, you know, we're looking at this. We're looking at the fact that the time is equal for both, so I'm gonna solve this one for time. Time is gonna be equal to the distance divided by the velocity. Get that by multiplying both sides by P and dividing both sides by V uh, teas overhere. Cancel these over here. Cancel. I end up with this equation here. Okay, so this equation is going to give me what this is going to say. This time is the same for both. And it is going to be equal to the distance of a man invited by the velocity of the man equal to the distance of the bear. Invited by the velocity of the bear, Uh, one myself in for. Well, here's what I'm looking for right here. I'm looking for Yeah, distance off the man, which is D So let's go ahead and solve this equation for the distance of the man. To do that, I'm gonna do pretty much the same thing is they did over here. I'm gonna multiply both sides by the velocity of the man on both sides, and that is going to cancel over here. So I end up with distance of the man, is equal to the distance of the bear, multiplied by the ratio off their velocities velocity of the man divided by the velocity of the bear. Okay, if you followed me to this point, we can go ahead. And now sit the two numbers in. So what is the distance off the man that is deep D Max? Because they have the same time. Okay, so, T Max is when time are equal. Okay, the distance of the bear is going to be equal to D plus 26. And then we have here the velocity of the man, which is four divided by the loss of the bear, which is six. Uh, this is wrong, actually. And the reason why it's wrong is because you have to multiply the D plus 26 times. The 46 not just 26. D plus 26 is the distance of the bear. All right, And then here, I'm gonna go ahead and just subtract or erase the fact that this is D Max because it's simply d in the problem. In order to solve this, I am going to take it on, rewrite it. So I'm going to say that Dean is equal D plus 26 times for sixths, and then I have a parentheses around this as well. Gonna multiply both sides by six and you end up with six D is equal to multiply Miss out. I ended up with four D plus 26 times, for which is 104 All right now, Miss attract for D from both sides and I end up with to D is equal to 104. Therefore, D, which is D Max when the time is equal, is going to be equal to 104 divided by two, which is 52 meters. I should put in here that assumes a magic door and we're done.

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