# Video: Calculating the Stopping Distance given the Reaction Time, Vehicle Speed, and Deceleration

The driver of a car traveling at 20 m/s has a reaction time of 1.2 s. The car’s brakes decelerate the car at 4.5 m/s² once they are activated. What is the car’s stopping distance to the nearest meter?

09:08

### Video Transcript

The driver of a car travelling at 20 meters per second has a reaction time of 1.2 seconds. The car’s brakes decelerate the car at 4.5 meters per second squared once they are activated. What is the car’s stopping distance to the nearest meter?

Okay, so in this question, we are considering a car that’s initially moving. And then it’s going to stop. So let’s say that this orange box is representing our car. Now, we’ve been told that this car is initially travelling at 20 meters per second. So let’s arbitrarily choose that the car was initially travelling towards the right at 20 meters per second. This is the initial velocity of the car which we will call 𝑢. Now what’s happening to this car is that is initially travelling at 20 meters per second. And then a driver spots a hazard.

So let’s say that this is the hazard that the driver spots. It’s a massive boulder in the way of the car. Now, let’s also say that the driver spots the hazard when the car is at this position here. And from that point on, the driver starts processing that they need to break. However, because the driver has a reaction time of 1.2 seconds, they do not step on the brakes instantly. In other words, the car still travels at 20 meters per second a short distance before the driver has even pressed the brakes.

So for this entire distance from here all the way to here, the driver along with the car has continued to travel at 20 meters per second while the driver is thinking about pressing the brakes. Now, this particular distance is known as the thinking distance which we will call 𝑑 subscript 𝑡. But then as soon as the car reaches this point, the driver does press the brakes. And it’s at that point that the car starts decelerating. In other words, from this point forward, the car is now accelerating in the direction opposite to that of its original motion at a rate of 4.5 meters per second squared.

Now, it’s at this point that we see that the acceleration that the car experiences due to its brakes is in the opposite direction to its initial velocity. And that’s, of course, because the car is decelerating. But because we’re dealing with the velocity of the car initially being in the opposite direction to the acceleration, it is a good idea to choose sign conventions. Let’s choose that anything towards the right is moving in the positive direction. And anything towards the left is moving in the negative direction. In other words then, the initial velocity of the car, which was towards the right, is positive 20 meters per second, whereas the acceleration or rather the deceleration of the car, which is towards the left, will be labelled negative 4.5 meters per second squared in our calculations.

Now, let’s call the acceleration of the car 𝑎. And then as soon as the car starts to decelerate at this point, we know that eventually the car is going to stop, hopefully before it reaches the hazard. So at this point, the car has stopped which means that the final velocity of the car, which we will call 𝑣, is zero meters per second because the car is now stationary. So the distance that the car travels over the period of time that it’s decelerating is known as the braking distance, which we we’ll call 𝑑 subscript 𝑏.

Now, the reason it’s called the braking distance is because this is the distance over which the brakes of the car have been applied. And by the way, we should also label that at the beginning of the braking distance, the velocity of the car was still 20 meters per second. Now, what we’re being asked to do is to find the car’s stopping distance. So let’s recall that the stopping distance of a car is defined as the thinking distance plus the braking distance. So let’s say that the stopping distance of the car we will call 𝑑 subscript 𝑠. And as we’ve already seen, thinking distance is 𝑑 subscript 𝑡. And braking distance is 𝑑 subscript 𝑏.

In other words then, the stopping distance is this total distance here. It’s the distance travelled by the car between when the driver first realises that there’s a hazard up ahead and when the car finally stops. So let’s write down over here that stopping distance is equal to thinking distance plus braking distance. And let’s go about working out the thinking distance and the braking distance individually.

Let’s start with the thinking distance. Now, as the car is moving from the point at which the driver realises they need to break to the point that the driver actually does press the brakes, as we said earlier the car continues to travel at 20 meters per second because the brakes haven’t been pressed yet. In other words then, the car is travelling at a constant velocity 𝑢 over this entire distance. And then we can recall that if an object is travelling at a constant velocity, let’s call that velocity 𝑣 subscript const, then that velocity is equal to the distance travelled by the object divided by the time taken for that object to travel that distance. And so in our particular case, the car is moving at a constant velocity of 𝑢. And this constant velocity is equal to the distance travelled, so that’s the thinking distance, divided by the time taken for that car to travel that distance.

Now as we know, the car actually travels this distance in a time of 1.2 seconds because that is the reaction time of the driver, in other words, the time taken between here, when the driver realises they need to brake, and here, when they actually press the brake. And so we know the value of 𝑢, 20 meters per second. And we know the value of 𝑡, the reaction time, 1.2 seconds. So we can rearrange this equation to solve for 𝑑 𝑡. We do this by multiplying both sides of the equation by the reaction time 𝑡. This way, it cancels on the right-hand side. And what we’re left with is that the reaction time of the driver multiplied by the constant speed at which the car is moving is equal to the thinking distance.

We can then say that the thinking distance is equal to the reaction time, 1.2 seconds, multiplied by the velocity of the car, 20 meters per second. And thinking about units very quickly, we’ll see that we’ve got seconds in the numerator here. And then we’ve got a per seconds, which means that the seconds units are going to cancel. And we’re going to be left with a unit of meters. Now, this makes sense because we’re calculating a thinking distance. So evaluating the right-hand side of this equation, we find that the thinking distance of the car is 24 meters.

So let’s write down that piece of information here. We’ve just worked out what 𝑑 𝑡 is. So in order to work out 𝑑 𝑠, all we now have to do is to work out the value of 𝑑 𝑏, the braking distance. In order to do that, we need to realise that, in this phase of the car’s motion, the car is no longer travelling at a constant velocity. In fact, it’s decelerating at a constant rate of 4.5 meters per second squared or, if you like, accelerating at negative 4.5 meters per second squared.

So what we do know about this phase of the car’s motion is that the initial velocity is 20 meters per second. And that velocity is positive cause it’s towards the right. We also know that the final velocity of the car is zero meters per second over here. And we also know the acceleration of the car, negative 4.5 meters per second squared, because remember the acceleration is towards the left, in other words, against the motion of the car itself. So if we know these three quantities, 𝑢, 𝑣, and 𝑎, and we’re trying to find out the braking distance 𝑑 𝑏, then to do this we need to recall one of the kinematic equations.

The specific equation we’re looking for is this one, the equation that tells us that the square of the final velocity is equal to the square of the initial velocity plus two multiplied by the acceleration of the object multiplied by the distance moved in a straight line by the object. But then in this situation, the distance moved in a straight line by the object, which in this case is the car, is actually the braking distance. And hence, we can replace 𝑑 𝑠 in the equation with 𝑑 𝑏, the braking distance.

We can keep all of the other quantities the same because we’ve called them 𝑢, 𝑣, and 𝑎. And of course, they mean the same things. So at this point, we can take our equation and rearrange to solve for 𝑑 𝑏. If we start by subtracting 𝑢 squared from both sides of the equation, what we’re left with is 𝑣 squared minus 𝑢 squared on one side and only two 𝑎𝑑 𝑏 on the right. Then we divide both sides of the equation by two 𝑎 to give us 𝑣 squared minus 𝑢 squared divided by two 𝑎 on the left and just 𝑑 𝑏 on the right. So now, all we have to do is to plug in some values.

When we do this, we find that we’ve got 𝑣 squared minus 𝑢 squared divided by two 𝑎. And then we can simplify the numerator first of all which becomes negative 400 meters squared per second squared, when we remember to square everything inside the parentheses from earlier, and the denominator becomes two multiplied by negative 4.5 meters per second squared which becomes negative nine meters per second squared. And then at this point, we see that the negative sign of the numerator and denominator cancel which is why it was so important for us to have accounted for the direction of the acceleration of the car. And if we think about units, we see that we’ve got meter squared per second squared in the numerator and meters per second squared in the denominator.

We’ve got the same powers of second. And they’re gonna cancel. But in the numerator, we have meters squared, whereas at the numerator, we just have meters. So only one power of meters in the numerator will cancel with the one power of meters in the denominator. And the final result is going to be in meters, which is great because once again, we’re finding a distance. So our distance is going to be 400 divided by nine meters. When we evaluate this, we find that it becomes 44.44 so on and so forth meters.

So let’s write that information up here. 𝑑 𝑏 is equal to 44.4 recurring meters. And at this point, we realise that we’re nearly there. We see that 𝑑 𝑠, the stopping distance which is what we’re trying to calculate, is equal to the thinking distance plus the braking distance. And so we can say that the stopping distance is equal to 24 meters plus 44.4 recurring meters. And that ends up being 68.4 recurring meters. However, we’ve not yet reached the final answer to our question.

Remember, we’ve been asked to find the car stopping distance to the nearest meter. So we need to round 68.4 meters to the nearest meter. To do that, we need to look at this value here after the decimal point. That value is a four which is less than five. And, therefore, this value, the eight, is going to stay the same. It’s not going to round up. And now, we’ve reached the answer to our question. The stopping distance of this car is 68 meters to the nearest meter.