You are driving to the grocery store at 20 m/s. You are 110 m fro...

Question

You are driving to the grocery store at 20 m/s. You are 110 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration. What magnitude braking acceleration will bring you to a stop exactly at the intersection?

Accepted Answer

Hey, everyone in this problem, a cyclist pedals at a constant velocity of 8m/s. He noticed a level crossing sign on the road 50 m away from his position. After one second, the cyclist decides to apply a constant deceleration so that he stops just before hitting the level crossing sign were asked to determine the applied deceleration. Now, we're told that this is a uniform or constant deceleration, but that means that we can use our um equation. They are uniformly accelerated motion equations or quadratic equations. So let's go ahead and write out the information we know. Now the cyclist is pedaling at a constant speed. And for one second, they're going to continue at the constant speed before they apply the deceleration. So this is a problem where we have to sort of stages to consider. Okay. So the first stage stage one is going to be before deceleration. Now, before deceleration again, we have a constant speed or constant velocity. And so we don't need to worry about five variables. We only have three variables to worry about V D and T. So the speed V one or the velocity it's going to be eight m/s. We don't know how far we travel, but we know that this stage lasts one second. So we have the philosophy, we have the time we can go ahead and find the distance. We're gonna hold off on that just for now. Because what we're really trying to find is a deceleration which happens in stage two. Okay. So if we don't need to find that distance, then let's not do the extra work. Let's look at stage two. First. Stage two is when we have deceleration, We have deceleration, we know that the initial speed it's going to be 8m/s. Okay. That's that constant velocity we're traveling with before we start to decelerate. So that's the initial speed before our deceleration. And this is going to be v naught to, we're going to use the subscript two to indicate stage two VF two. The final speed is going to be 0m per second because the cyclist is going to come to arrest or come to a stop before hitting the crossing side. We don't know what the acceleration A two is. That's what we're looking for. Okay. That deceleration. So we expect our acceleration to be negative indicating a deceleration. We don't know how far we travel and we don't know how long this takes. Alright, so we look at our stage to information we have two values that we know we not in V F. We want to find the acceleration. In order to be able to use our um equations, we need three known values. Well, we can figure out the distance D one, can we relate that to stage two, we know that the Crossing sign is 50 m away and we're going to stop right before what that means is the total distance D one plus de two in the distance we travel on stage one plus stage two is equal to 50 m. So we can write this distance D two as 50 m minus the distance D one. So we can go ahead and calculate the distance D1 and use that in our stage two equation. So we'll call for constant speed. We have V is equal to D over T velocity is equal to distance divided by time. Substituting in the information we have, we have eight m per second is equal to the distance traveled in stage one, divided by the time one second. And so the distance traveled during stage 18 m per second times one second gives us eight m. So that is the distance we travel that cyclist travels in that 1st 2nd before they apply the deceleration. This tells us that the distance traveled in stage two, it's going to be 50 m minus the eight m we travel in stage one For 42 m. All right. Now, if we look at stage two, we have three known values. We have the not, we have V F and we have the distance or the displacement, we want to find the acceleration. So we're going to choose the equation that has those four variables. The one that doesn't have the time T in it because we don't know the time T and that's not what we're looking for. And the equation will be the following V F squared. And it's VF two squared is equal to V, not two squared Plus two multiplied by the acceleration multiplied by D two. On the left hand side, we have zero m squared per second squared. On the right hand side, we get eight m/s, all squared last two multiplied by acceleration. A two multiplied by 42 m. So we have zero is equal to m squared per second squared plus 84 m multiplied by A two. This tells us that A two. If we rearrange is going to be equal to negative m squared per second squared divided by 84 m, we have meter squared per second squared divided by meters. This is going to leave us with a unit of meters per second squared, which is what we want for acceleration. And we get a value of negative 0. meters per second squared. Now, if we look at our answer choices, we have answer choice is a negative 8m/s squared. B negative four m per second squared C negative 0.76 m per second squared and D negative 0.64 m per second squared. We can see that the answer choices use two significant digits. And so if we take our answer of negative 0.762 m per second, we write this with two significant digits. We see that this corresponds with answer choice C -0.76 m/s squared. Thanks everyone for watching. I hope this video helped see you in the next one.

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You are driving to the grocery store at 20 m/s. You are 110 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration. What magnitude braking acceleration will bring you to a stop exactly at the intersection?

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