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. b. What is her speed as she passes him?

Video transcript

Hey, everyone in this problem, a runner is at rest at the start line of an athletic field and is passed by his trainer who is moving at a constant speed of 4.5 m/s on his bike. When his trainer passes the start line, the runner begins to accelerate, increasing his speed by applying constant acceleration of 0.75 m per second squared. And were asked to determine the speed of the runner when he catches his trainer. The answer choices we have are a three m per second, be four m per second. C six m per second and D nine m per second. And so this is a problem with quadratic equations. Can we have emotions? We're gonna use our kid a Matic or you am equations, but we have two things to consider. We have the runner and we have the trainer. So let's start with the trainer. The trainer is moving with constant speed. So the only variables we have to worry about for the trainer are the speed, the distance and the time. So we have the velocity Okay. And we're gonna subscript with T for the trainer, we know that, that's 4.5 m/s. Now, the distance this trainer travels, we're gonna call this distance D C okay. And the sea is gonna indicate patch, okay. So this is the distance the trainer travels before the catch anyone. The runner catches him. Similarly, for t we're gonna say T C the time that's elapsed between when they start until the runner catches the trainer. So this is for the trainer. Now, what about the, we're told that the runner is at rest at the start line. Okay. So the initial speed of the runner be not, is going to be equal to 0m/s. The final speed of the runner, we don't know, but that's what we're trying to find. We're trying to find the speed of the runner when he catches the trainer. So we don't know the final speed, The acceleration of the runner we're told is 0.75 m/s squared. Now the distance the runner travels, this is gonna be the same as the distance the trainer travels when the runner catches a trainer. Okay. Once the runner catches a trainer, they've traveled the same distance and the same amount of time has passed. So D C and T C are gonna be the same between the trainers equation and between the runners equation. Alright, so we want to find VF we only have two known quantities we know V not and we know the acceleration A Okay. So that's not enough to solve for V F directly using just the runners information. Okay. We need three known quantities. We only have two. So in order to find via, we're gonna need a soul for either DC or TC, the distance or the time traveled. How can we find those? Well, with the trainer, we only have one piece of information. So that's also not enough to solve for the distance or the time. But what we'll notice is that we have 21 notes in two equations. Okay. So we have two unknowns for the trainer equation. We can write two unknowns in the runner equation. And we know that if we have two equations with two unknowns, we can substitute them into each other and solve for those unknowns. So let's start by writing an equation for the trainer and for the runner. So starting with the trainer again, they have a constant speed. So the equation we're going to use here is just velocity equals distance over time. So we have V T Equal to DC over TC. We know VT is 4.5 m/s. Yes, we have 4.5 m per second is equal to D C over T C. And let's just isolate D C. So we can multiply TC up, we have D C is equal to 4.5 m per second times TC. And we're gonna call this equation one. Now, moving on to the runner. We need to include D C and T C in our equation in order to have those same two unknowns as our trainer equation so that we can actually solve from one of these variables. Okay. We know we not, we know a so what we need to do is choose the equation without the final speed V F. Okay. That seems a little bit backwards because that's the value we're looking for. But at this step, we need to include D C and T C and R two known values. So we're gonna use the following equation or the distance T C is equal to the initial speed. V naught times the time T C plus one half the acceleration a times the time T C squared we can substitute in what we know. So we have the D C is equal to V naught is zero. So V nought times T C that term just goes to zero. We're left with one half times the acceleration, 0.75 m per second squared times T C squared. Alright. So let's call this equation to And we're gonna substitute equation one into equation two. So we substitute equation one into equation two. Equation one is D C is equal to 4.5 m per second times T C. So we put that into the left hand side where we have DESEAN equation two and we get 4.5 meters per second times T C is equal to one half times 0.75 is going to be 0. meters per second squared times T C squared. And we want to solve for T C. So we're gonna move all over TC terms to the right hand side and have zero on the left hand side, you can see that we have a T C squared term and just a T C term. So this is like solving a quadratic function. So when we do that, we get zero is equal to 0. m per second squared times T C squared minus 4.5 m per second times T C. And we don't have a constant term here. So we can factor out one of these TCS, the factor out TC and we are left with 0.375 m per second squared A T C squared divided by T C. We're still left with one T C here -4.5 m/s. Now we have two factors multiplied together. So this equation is going to equal zero if either of those factors are zero. So the first factor is just T C and so we get that T C is equal to zero is a solution. And that makes sense because at time zero, that initial time we're starting at the trainer is just passing the runner, okay. So they are at the same position at time. Zero. That makes sense in the other cases when we have this whole factor equal to zero. So we have 0.375 m per second squared times T C minus 4. m per second, equal to zero. This tells us that TC is equal to 4. meters per second, divided by 0.375 m per second. Squared. This is going to leave us with a unit of seconds, which is what we want for time And we get 12 seconds. And so the time, the second time that they're at the same position, which is going to be the time that the runner catches the trainer is going to be 12 seconds. Alright, that's great. But remember we're looking for the speed of the runner when he catches the trainer. So let's go back up and look at our equations. We have not, we want to find VF we have a and now we also have TIT is equal to 12 m/s Or sorry, not 12 m/s, just 12 seconds. Alright. Now that we have three known values, we can find VF that we've been looking for. We're going to use the equation that doesn't include the distance because we don't have information about that and that's not what we're interested in. So let's do that and sell for Viet. So the equation we're going to use is V F the final speed is equal to v naught plus a times the time of the catch. So that final speed when the runner catches a trainer is going to be equal to the initial speed is just zero. So that goes to zero, The acceleration is 0.75 m/s squared. And the time That the runner catches a trainer that we found was seconds 0.75 m per second squared times seconds, that's going to give us a unit of meter per second that we want for speed And we get nine m/s. And that is going to be the speed of the runner when he catches the trainer. If we go back up to our answer choices, we see that that corresponds with answer choice D nine m per second. Thanks everyone for watching. I hope this video helped see you in the next one.