David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.0 m/s² at the instant when David passes.

a. How far does Tina drive before passing David?

Question

David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.0 m/s² at the instant when David passes.

a. How far does Tina drive before passing David?

Accepted Answer

Hey, everyone in this problem, we have a police officer at rest on a sidewalk is passed by a thief running at a steady speed of 3. m/s. Just as a thief exceeds him. The police officer accelerates at a rate of one m per second squared. And we're asked to determine the distance the police officer travels before catching the thief. The possible answer choices are a 3.5 m. B 12.3 m C 24. m and D 28 m. Now we're giving information about the thief and the police officer. So let's start with the thief. So we're told that they're running at a steady speed. If they're in a steady speed, they don't have acceleration. The only variables we need to worry about are the velocity, the distance and the time. And I've used subscript T to indicate thief. Now we're told they have a steady speed of 3.5 m per second. So V T is going to be 3.5 m per second. We're not told about the distance and we're not told about the time. So let's move over to the police. Now, the initial speed of the police is going to be 0m/s. They're standing at rest on the sidewalk and then they start to run after the thief passes them. We don't know the final speed of the police officer. We know that they accelerate At a rate of one m per second squared. We don't know the distance they travel, but this is what we want to find. We also don't know the time but let's think about the relationship between the thief and the police officer. Mhm The movement that we're considering starts when they're at the same position, the police officer starts as soon as the thief passes him. Okay. So the distance the thief in the police officer travel when the police officer catches him is going to be the same because they start at the same position and they're going to end at the same position. So the distance is going to be the same. So this value D T is going to be equal to the same DP. It's for D P, same thing, this is equal to D T and this is what we're looking for. Now, the time is gonna be the same thing. They start at the same time and then when the police officer catches a thief, it's at another time and the time point for the thief and the police officer are gonna be the same. And so T T is equal to T P and vice versa. Alright, so we want to find the distance D P, we can't do that directly from the information we have about police officers because we only have two known values. We have the initial speed and the acceleration to use RU AM or Kinnah Matic equations to solve for D P. We need three known values. Okay. So we need to figure out one more of these unknowns first before we can solve for D P using just the information with the police officer. However, we know this relationship between the distance and the time for the thief and the police officer, they're equivalent, they're the same. So what that gives us is two unknown values D and T, but we can have them into equations, the thief equation and the police equation, two unknowns into equations we can solve using substitution. So let's do that. Let's start with the equation for the thief. And again, they're moving at a constant speed. So the equation we use is that the velocity is equal to the distance over the time we're gonna substitute these T values in for P values for the distance in the time. We know that this speed is 3.5 m per second. And that's going to equal DP divided by T P. And let's just write this in terms of D P. And so we get that D P is equal to 3.5 m/s times TP And we're gonna call this equation one. And then we're going to do the same for the police officer. So for the police officer, we want an equation that doesn't include this final velocity. We don't know the final velocity or the final speed. That's not what we're trying to find. And it's not one of the values that we can relate to the thief. So we're gonna choose the equation without that. And that's the following that the distance D P is equal to the initial speed. V, not P times the time T P plus one half the acceleration A P times the time T P squared filling in what we know we have D P is equal to the initial speed is zero m per second. So V nought times T is going to be zero and we have one half times the acceleration, one m per second squared times a time T P squared. We're gonna call this equation to and now we're in a substitute one equation into the other because we have two equations with the same two unknowns D P and T P. We're gonna substitute equation one into equation two. And when we do that, we're gonna have the following. We're gonna have 3.5 m per second times T P is equal to 0.5 m per second squared. Okay. We have one half times one m per second squared of 0.5 m per second squared times T P squared. Now, this is like a quadratic function of T P. We have T P squared term, we have a T P term. So let's move everything to the right hand side and have this equal to zero so that we can solve for T P. So we get zero is equal to 0.5 m per second squared T P squared minus 3.5 m per second times T P. We can factor out T P because we don't have a constant term. So we have T P. In both terms, we get T P times 0.5 m per second squared times T P minus 3.5 m per second. And now we have two factors multiplied together. We want them to equal to zero. If either factor is equal to zero, then the product will be equal to zero like the left hand side and we will satisfy this equation. So the first factor is just T P. So T P equals zero. This equation is satisfied. And that makes sense. Okay. At that initial time, when time is zero, the thief is just passing the police officer while the police officers at rest on the sidewalk before the police officer starts chasing him. Okay. So at that initial time there at the same position, that makes sense. Okay. What we're worried about is when the police officer catches the thief after he starts chasing and that's gonna come in the second part. So if we have this second part equal to zero, we have 0.5 m per second squared times T P minus 3.5 m per second is equal to zero. This gives us A T P is equal to 3.5 m per second, divided by 0.5 m per second squared. And we get a time of seven seconds. So the police officer is going to catch the thief after seven seconds. So we have the time. Now, we can substitute that back to either of our equation to find the distance. And the distance is what we're looking for. And we had equation one, we had equation two, equation one is a little bit simpler. So let's use equation one, we're gonna sub that time we found in into equation one. And again, this is the time when the police officer catches the thief. And so using this time is going to give us the distance when the police officer catches that thief. So DP is equal to 3.5 meters per second times the time of seven seconds. And we get a distance of 24. m, okay. And so the police officer is going to catch the thief in seven seconds and he is going to travel 24.5 m before doing. So we go back up to our answer choices. We see that that corresponds with answer choice C24. m. Thanks everyone for watching. I hope this video helped see you in the next one.

David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.0 m/s² at the instant when David passes. a. How far does Tina drive before passing David?

Verified video answer for a similar problem:

Key Concepts

Relative Motion

Kinematics Equations

Time of Travel