A cheetah spots a Thomson's gazelle, its preferred prey, and leap...

Question

A cheetah spots a Thomson's gazelle, its preferred prey, and leaps into action, quickly accelerating to its top speed of 30 m/s, the highest of any land animal. However, a cheetah can maintain this extreme speed for only 15 s before having to let up. The cheetah is 170 m from the gazelle as it reaches top speed, and the gazelle sees the cheetah at just this instant. With negligible reaction time, the gazelle heads directly away from the cheetah, accelerating at 4.6 m/s² for 5.0 s, then running at constant speed. Does the gazelle escape? If so, by what distance is the gazelle in front when the cheetah gives up?

Accepted Answer

Hey, everyone in this problem, we have a training situation with police officers. I'm during a thief arrest, police officer notices a thief robbing a car. They run in the direction of the thief attaining a constant speed of 6.5 m/s. Unfortunately, the police officer could only maintain that maximum speed for 25 seconds before having to stop the instant. The police officer is at his maximum speed. The thief is a warrant And at that instance, the police officer and the thief are 75 m apart. The deed starts to run accelerating at a constant rate of two m per second squared for three seconds. And then he maintains his speed and were asked, does this thief end up in jail? Okay. So does the police officer catch the thief? And if not, how far ahead is a thief when the police officer stops were given four answer choices. The first is, yes, the police officer catches a thief. It takes eight seconds. B is yes, the police officer catches a thief after 23 seconds answer choice C is no, the police officer stops. The thief is 8.5 m ahead and D is know when the police officer stops, the thief is 53.5 m ahead. Alright, so that was a lot of information given let's just draw a little picture of the initial situation. So we have our police officer, we're going to say that he is running to the right. Okay. We're gonna take right to be positive X direction And he's moving at a constant speed of 6.5 m/s. Now we're gonna take it at the instant that he achieves a speed, which is when he warns the thief. So we have our thief in his initial speed is zero. He's not moving just yet, but he's going to start moving and the distance separating the two of them is 75 m and this thief has some acceleration to the right This positive acceleration of two m/s squared. Alright, so let's try to write out the details of the variables were given. Let's start with the police officer. Let me actually go up and make use of this room here, the police officer And then moving into constant speed. So we only have three variables that we need to consider. We have the speed of the police officer VP which we know is 6.5 meters per second. We know that the police officer can maintain this speed for seconds. We don't know how far the police officer runs in those 25 seconds. We're trying to figure out whether the police officer ends up catching the thief. So let's figure out how far both the police officer and the thief run and then we can compare the distances to see if the police officer catches them. So let's go ahead and find delta X P. So recall when we have constant speed, we have at the speed is equal to the distance over the time, which tells us that the distance delta X P we're looking for is going to be equal to the speed V P multiplied by the time T P, This is equal to 6.5 m/s times 25 seconds for a distance of 162.5 m. Alright. So from this initial point that we've drawn, the police officer is going to run 162. m before coming to a stop. Let's move over to the thief now and see how far the thief brand in order to see if the police officer will catch him. So for the thief, we have to consider kind of two stages of motion. Okay. We have the initial stage stage one where the thief is accelerating and then we have a second stage where he's moving at a constant speed. Now, in this initial stage, we know that he's starting from rest. So we not we're gonna call it be not one to indicate stage one of the thief And this is zero m/s. We don't know what the final speed will be. We know that he accelerates At two m/s squared. We know that they spend three seconds accelerating, oops delta X and delta T. Alright. So we have three known values here so we can use our you am or Kinnah Matic equations to solve for the others. Okay. We definitely want to solve for delta X one because that's part of the distance of the thief travels. Let's move to stage 2 1st before we solve for anything in stage one and see what we have. So in stage two, we have constant speed. So we just have three variables to worry about. So we have V two, which is gonna be the speed of the thief in stage two and we don't know this value. However, we know that the thief accelerates to a maximum speed and then maintains that speed. So the final speed from stage one Is going to be the speed that we're traveling in stage two. So V two is going to be equal to VF- one. We want to find Delta X two and the time T two. What's the time we know that 25 seconds elapses okay. In total, The police officer maintains at max speed for 25 seconds and so the thief is running away for 25 seconds. We spend three seconds in stage one. So in stage two, the thief is going to spend 25 seconds minus the three seconds spent accelerating for a total of 22 seconds. Okay in this constant speed phase. Now again, what we're looking for is how the distance or the displacement of this thief. So we want to take the displacement and we're going to say delta X T of the thief. Let me go back up to our picture for a second. We want to take the displacement from the initial position of the police officer instead of the initial position of the thief so that we can compare the displacement of the police officer in the thief Against each other. Okay, we'll call it the thief started 75 m ahead of the police officer. So they already have a head start. And so the distance or the displacement of this, the relative to that initial position of the police officer is gonna be the initial position plus the amount they travel during stage one, the acceleration phase plus the amount they travel during phase two. That constant speed phase. We know delta X I is 75 m. So we need to find delta X one and delta X two. Now delta X one, we've already said we can use our Kinnah Matic equations to solve for, for delta X two. We only know T two, we don't know V two. However, we know V two is equal to V F one. So we can use our Kinnah Matic equations in stage one to find V F one and then we can solve for delta X two. Okay. So we have a way to get all these values we want, we're just gonna have to go step by step. So let's start by finding delta X one. So we're going to choose the Kinnah Matic equation that includes the variables we know V not A T and the one we're trying to find delta X. So we're going to exclude VF and we have the following delta X one Is equal to V, not one multiplied by T one plus one half multiplied by a one multiplied by T one squared. This gives us delta X one V, not a zero. So this first term goes to zero, we got one half multiplied by two m per second squared, Multiplied by three seconds squared. Okay. So one half times two, that gives us one m per second squared multiplied by nine seconds squared. This is going to give us a value of nine m. So during the acceleration phase of the Thief's motion, we're gonna move nine m. Now, we need to figure out how far they move during that constant stage, constant speed stage. So again, we need to find VF- one in order to find delta. Okay, to find VF1, we're going to choose any of our cinematic equations. Can we know all of the other four variables? Now, so let's just choose the one that we think is going to be the simplest VF one is equal to V not one, A one multiplied by T one, the initial speed V not is zero. So that term goes away, We have the acceleration two m/s squared Multiplied by three seconds. The time This is going to give us a speed of six m/s. Alright, so now we know V F one is six m per second. If we look at our stage two Information that tells us that V two is also going to be six m/s. So now we can calculate Delta X two. Let's do that right on the side here beside our stage to information. So again, this is constant speed. So we know that the speed V two is going to be equal to the distance. Delta X two divided by the time T two. So Delta X two will be equal to V two multiplied by T two. This is equal to six m/s times seconds For a total distance traveled of 132 m. All right. So the thief is gonna travel nine m while they're accelerating 132 m while they're in that constant speed face. So they don't actually run as far as a police officer in that amount of time. However, they had a head start, okay. So remember let's calculate that total distance okay relative to the initial position of the police officer. So we have 75 m that initial distance separating the two of them plus nine m, the distance traveled in the acceleration phase plus 100 and 32 m for a total of 216 m. Okay. So the Thief's final position is gonna be 216 m away from the initial position of the police officer. Well, recall that the police officer only moved 162.5 m. So the police officer is not going to catch the thief and the difference in their displacements delta X T minus delta X P is equal to 216 m -162.5 m, Which is 53.5 m. And so the police officer is going to be 53.5 m behind the thief when he stops running. Alright, so if we go back up to our answer choices, okay. We found that the police officer is not going to catch the thief. So we're looking at option C or D, We found that the thief is going to be 53.5 m ahead. And so we have answer choice. D thanks everyone for watching. I hope this video helped you in the next one.

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