Practice: Two horny teenagers run towards each another. The guy, initially at the bar, runs with 5 m/s. The girl runs from 60 m away from the bar with 4 m/s. How long does it take for them to meet?

Subjects

Sections | |||
---|---|---|---|

Intro to Motion (Kinematics) | 67 mins | 0 completed | Learn |

Motion with Multiple Parts | 53 mins | 0 completed | Learn |

Meet/Catch Problems | 23 mins | 0 completed | Learn |

Vertical Motion | 48 mins | 0 completed | Learn |

Concept #1: Intro to Meet/Catch Problems

**Transcript**

Hey guys, I want to talk about a very specific type of motion problem that I call the meet catch problems, these are very usual in physics one and in physics tests and I want to show you how this works first I'm going to start with a very simple example and then I'm going to talk about it, so let's get started. Alright so meet catch sometimes I might refer to these as chase problems one animal is chasing the other but they eventually one catches the other. So very quick example here 300 mile trip so your trip has a length of 300 miles and your friends are ahead of you by 50 miles you guessed it their speed is 55 so velocity of your friends is 55 it's average. So their acceleration is 0 and you want to know how fast do you have to go so that you catch up to them. So what is the velocity that you have to have average velocity so that you catch up to them in other words so you arrive at the same time which means that delta T of your friends is the same as delta T of you, and this is what's going to happen in these problems these meet catch problems your time will be the same as their time or whatever two objects are moving. So if you want to draw this real quick your friends are going from here this is 50 miles your friends are going from here to here right this whole thing is 300 so this is 250, while you're going from here the very beginning all the way over here. So you're actually covering 300 miles so you don't start from the same place, the time starts counting here and the time ends counting here but the idea is that it's the same time so delta T equals delta T, it's the same times. The question is how fast are you moving? So for this interval they're moving at 55 and I want to know how fast are you moving if you go straight for the answer which is probably what you should do you see that the average velocity equation is delta X over delta T you know your delta X, your delta X is 300 it's how much you're travelling but we don't have the time. So it's one of those where you're stuck but there's another situation that is related to this one where you can find the time the time for you is the same as the time for them and I have enough information to figure out how long it takes for them to get there so that's what we have to do. So I have to find this time here so we are going to go to your friends and figure out what is there time the only equation again is V over delta X over delta T, so time delta T is delta X over V. Now delta X is 250 not 300 and their velocity is 55 let me talk about units real quick because this is 250 miles and this is 55 miles per hour so these two cancel and you have hours obviously it's time so it's going to be hours and you have 4.55 hours. An important point to make here is that you don't have to convert anything everything in miles or hours and you want to know the answer in meters per hour so you can just keep everything as it is all these units are consistent so I can just plug in that 4.55 hours here and I get the answer which is 65.9 miles per hour. You have to go faster than they are travelling that makes sense because you're trying to catch up right so that's a very simple one that we could've solved without me even talking about this anymore giving you any more details about this new type of problem. But let's get into that now because we're going to have some more complicated stuff. So when two objects meet the basic idea is that they are at the same place at the same time and that's pretty obvious. But more specifically this means that the X final of A is the same as X final of B and the time of A is the same as time of B, they have the same time.

So how are we going to solve this there 4 steps. The first one is we're going to write their position equations for a each object and the position equation I will remind you it's the third equation but I'm going to draw out a little I'm going to write a little different I'm going to say X equals X initial plus B initial of T plus half of A T squared. So it's this except that I expanded that delta X into X minus X initial moved the X initial to the other side alright so it's this one that's the position equation you write this for both objects then you set them equal to each other you equate their position equations this is the most important part of the instruction why because if you remember equate position equations it's obvious that for you to set them equal to each other you obviously have to write them first right so if you remember step two you're going to remember step one and if you remember step two you're going to remember step three when you set these two equations equal to each other as I'll show you soon the only variable you have left is time and then the only thing left to do in that equations is to solve for time so as long as you remember step two you should remember everything else. When to solve for time if you're looking for the distance at which they meet or one catches the other all we have to do is plug time back into the position equations. So better than talking about this is to show this so let me show you how this works. So a car starts from rest, so car has the initial velocity of 0 and moves with a constant 8 meters per second to the right so the acceleration of the car is 8 meters per second at the same time a truck moving with a constant 30 so V average or V constant equals 30 which means the acceleration is 0 passes the car are they going to meet again? Yes, No If not why not. So do they meet again the answer is yes and it doesn't ask why do they meet again but basically the acceleration of car is greater than the acceleration of truck. This is 8 this is 0 so the car eventually catches up, so let me draw this real quick. This is the car moving with initial velocity of 0 acceleration of 8 and this is the truck the truck was already moving so the truck kind of comes from behind and then crosses over here alright so the initial velocity of the truck which is a constant velocity for the truck is 30 the acceleration is 0. The idea is that the time starts counting here because the car starts moving when the truck passes the car so this is my X equals 0 and the time is 0 for both of these. So what we have to do is write the position equation remember the position equation is X equals X initial plus V o T plus half of A T squared this is the expanded position equation this is your final velocity, the final position of the car is the initial position they both start in the same place so we call that 0. The initial velocity of the car is 0 as well and the car does have acceleration of 8.

I don't have time right and you should expect that you don't have time in these equations because you're trying to figure out when do they meet so you're looking for that time and eventually you're going to get that. So as you're plugging these numbers in in these specific types of problems you're not going to have time, so if you clean this up a little bit it is 4 T squared I'm going to put a box around it because we're done here. Now X truck the truck also starts at 0 but it has no initial velocity. It's a constant velocity so it is 30 T again you don't have T so you can just write 30 T and this truck has no acceleration so this is 0 so you get 30 T. That's the first step you write the position equation the second step is to set them equal to each other so if I equate these two equations I get 4 T squared equals 30 T. The easiest way the only way that I know to solve this is to cancel the T's so I can solve for T. T equals 30 over 4 which is 7.5 and this is seconds because the acceleration was in meters per second squared the time was in meters per second squared SI units so the time is just 7.5 seconds so that's how you can know these problems you equate their position equations once you do that notice that the only variable is T. That's why I said that the third step up here is an obvious one once you've written their equations right. So I got that time and that answers when they meet so they meet at 7.5 seconds. The next question is where do they meet in other words we're looking for X. Now I can plug this number into either one of these two equations right here and that's because they are the same at the same time right so you can plug this into either one of them. So it says that you can plug that into either position equation right and I can if you want to do this you could see that for 4, 7.5 T that's X car would give you 225 and we could do or X equals 30 times 7.5 that also gives you 225. So that's your final answer for where they meet. So you're just going to plug it into the simplest one of your position equations I want to answer the extra question there which is to find the car's final speed. So for this I can just use one of the motion equations one of the three motion equations I got them up here, the first one works the final velocity of the car is the initial loss of the car which is 0 plus acceleration time the car accelerates with 8 and it's been 7.5 seconds so we do this and we get 60 meters per second. This is for the answer for the bonus question and that's the final velocity of the car. Anyway so that's basically how you're going to do these you write the position equation set them equal to each other to find your time and then you're going to plug in that time back into either one of the position equations to figure out where they meet and sometimes they're going to ask you how fast was one of the two objects right, they're not going to ask how fast the truck was going because the truck is a constant 30 but the car is changing its velocity so it makes sense to ask for the cars velocity at the end, so that's it for this one.

Concept #2: Meet/Catch Problems (Practice Intro)

**Transcript**

Hey guys, I now want you to try two meet and chase practice problems. Remember you're supposed to write their position equations set them equal to each other and find the time let's give it a shot.

Practice: Two horny teenagers run towards each another. The guy, initially at the bar, runs with 5 m/s. The girl runs from 60 m away from the bar with 4 m/s. How long does it take for them to meet?

Practice: In a race, your friend takes off with 3 m/s^2. You take off 2 seconds later with 4 m/s^2. How far into the race will you catch up to him?

0 of 4 completed

A motorist traveling at a constant velocity of 20 m/s passes a school-crossing corner where the speed limit is 10 m/s (about 22 mi/h). A police officer on a motorcycle stopped at the corner starts off in pursuit with constant acceleration of 2.5 m/s2. What is the officer's speed when she catches up with the car?

A motorist traveling at a constant velocity of 20 m/s passes a school-crossing corner where the speed limit is 10 m/s (about 22 mi/h). A police officer on a motorcycle stopped at the corner starts off in pursuit with constant acceleration of 2.5 m/s2. What is the total distance the officer has traveled at the point he catches the motorist?

A 1/4 mile drag race is going to occur between a Porsche GT3 and a Ferrari F430. The Porsche is the faster car, able to accelerate across a 1/4 mile at about 6.5 m/s2, which the Ferrari can only accelerate across the same stretch at about 5.8 m/s2. At the start of the drag race, though, the driver of the Ferrari gets off the line faster than the Porsche, by about 0.5 s. Is this enough of a head start for the Ferrari to win? Note that 1/4 mile is about 400 m.

A car capable of a constant acceleration of 2.64 m/s2 is stopped at a traffic light. When the light turns green, the car starts from rest with this acceleration. At the very same moment, a truck traveling with constant velocity 17.3 m/s passes the car. As the car’s velocity increases, it will eventually move faster than the truck and later overtake it. How far from the light will the car catch up with the truck?
1. 171.127
2. 43.6985
3. 116.654
4. 131.742
5. 210.032
6. 162.959
7. 259.102
8. 53.2193
9. 226.735
10. 323.646

Two cars, an Acura and a Bentley, are initially at rest a distance L from each other. At time t = 0 both cars starts accelerating toward each other with constant acceleration aA and aB respectively.
a) At what time does the crash happen?
b) Find the distance traveled by the Bentley before the two cars crash into each other.
c) What is the relative speed of the Bentley, with respect to the Acura, just before the crash?
Write your results in terms of L, a A, and aB. Remember to check the dimensions/units for each answer.

Two locomotives approach each other on parallel tracks. Each has a speed of exttip{v}{v} 85 km/h with respect to the ground. If they are initially exttip{L}{L} 7.0 km apart, how long will it be before they reach each other?

A top fuel dragster and a Bugatti Veyron race in a short straight line drag race. At the start of the race the top fuelers is standing still waiting for the Bugatti to show up. The race starts when the Bugatti crosses the starting line at its top speed of 400 km per hour. At that instant, the top fuel driver accelerates to 4.5 gs and continues to do so for the entirety of the race. How long will it take the top fueler catch the Bugatti?

Suppose the elevator starts from rest and maintains a constant upward acceleration of 4.10 m/s2. A bolt in the elevator ceiling 3.10 m above the elevator floor works loose and falls. It falls out the instant the elevator begins to move. (a) How long does it take for the bolt to reach the floor of the elevator? (b) Just as it reaches the floor, how fast is the bolt moving according to an observer in the elevator? (c) According to an observer in the elevator, how far has the bolt traveled between the ceiling and floor of the elevator? (d) Just as it reaches the floor, how fast is the bolt moving according to an observer standing on the floor landings of the building? (e) According to an observer standing on the floor landings of the building, how far has the bolt traveled between the ceiling and floor of the elevator?

An inquisitive physics student and mountain climber climbs a 50.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.00 m/s. (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously? (c) What is the speed of each stone at the instant the two stones hit the water?

While driving on the highway at a constant speed v0, significantly above the speed limit, you pass in front of a parked police car without noticing it. After a few seconds, during which time you have moved a distance d, the police car starts chasing you with a constant acceleration a0. Give your answers in terms of v 0, d, and a0. Assuming that you keep moving at a constant speed, and that the highway can be considered straight:[a] How long will it take for the police car to reach you?[b] How far has the police car traveled when it reaches you?[c] What is the speed of the police car when it reaches you?

An unmarked police car traveling a constant 95 km/h is passed by a speeder traveling 135 km/h. Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car’s acceleration is 2.00 m/s2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)?

Cars A and B are racing each other along the same straight road in the following manner. Car A has a head start and is a distance DA beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed vA. Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed vB, which is greater than vA. 1. How long after Car B started the race will Car B catch up with Car A? 2. How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities. (You may use tcatch as well.)

Enter your friends' email addresses to invite them:

We invited your friends!

Join **thousands** of students and gain free access to **55 hours** of Physics videos that follow the topics **your textbook** covers.