Video Transcript
There are two identical vehicles on two different roads under the same conditions. The drivers of both vehicles encounter an obstacle, suddenly brake, and then come to a complete stop. Driver one has an initial speed twice that of driver two. How, if at all, will the braking distances of the two vehicles be different? (A) Driver one will have twice the braking distance of driver two because braking distance is proportional to initial velocity. (B) Driver one will have four times the braking distance of driver two because braking distance is proportional to the square of initial velocity. (C) Driver one will have half the braking distance of driver two because braking distance is inversely proportional to initial velocity. (D) They will be the same because braking distance doesn’t depend on initial velocity.
In this situation, we have two identical cars driving along two different roads. The roads are of the same conditions. And we’re told that the speed of driver one, we’ll call that speed 𝑣 one, is twice that of driver two, what we’ll call 𝑣 two. Under these conditions, we’re told that both of these drivers encounter an obstacle in the road. Say that it’s this fallen-down tree. Both drivers, as soon as they see the obstacle, press down hard on the brake to bring their vehicle to a stop.
Our four answer options all describe what’s called the braking distance of these two drivers. The idea with braking distance is this. Say that at this instant we’ve shown here, light coming from the obstacle on the road reaches the eyes of both of our drivers. Both drivers will need to process this information and then send a signal down to their foot to press down on the brake pedal. This takes time, and during that time the cars continue to move ahead.
Let’s say that by the time driver one begins to press down the brake pedal, car one is here. And by the time driver two begins to press down the brake pedal, car two is here. At this instant, the speeds of the two cars haven’t changed. The first vehicle still has speed 𝑣 one, and the second still has speed 𝑣 two. However, when the drivers start to apply the brakes, the vehicles do begin to slow down. Because these vehicles are identical, we’ll say that once braking begins, they both experience an acceleration 𝑎 to the left, as we’ve drawn it. This acceleration is constant as the cars slow down and come to a stop.
While these vehicles are accelerating, while they’re coming to a rest, they cover a distance called the braking distance. We would like to compare the braking distance experienced by driver one to that experienced by driver two. Because both vehicles experience the same constant acceleration, we can think of this situation in terms of the kinematic equations of motion. We recall that these equations apply whenever the acceleration experienced by some object is constant.
One of the four kinematic equations of motion says that the final velocity of an object squared is equal to the initial velocity of that object squared plus two times that object’s acceleration multiplied by its displacement 𝑠. In our situation, this displacement 𝑠 is the braking distance of a given vehicle. It’s how far the car will travel while it’s being accelerated or decelerated to a stop.
Let’s also note that for both of our cars, since they come to a stop, the final velocity of each one is zero. For either one of our cars then, we can write this kinematic equation as zero is equal to the initial speed of the car squared plus two times the acceleration experienced by the car while it brakes multiplied by its braking distance 𝑠. We can rearrange this equation so that the braking distance is the subject as follows. First, we subtract 𝑣 sub i squared from both sides. That cancels 𝑣 sub i squared out entirely on the right. Then, we divide both sides of the remaining equation by two times 𝑎. This causes both two and 𝑎 to cancel out on the right. We find then that the braking distance experienced by either one of the cars equals negative that car’s initial speed 𝑣 sub i squared divided by two times 𝑎. As we’ve seen, 𝑎 is the acceleration each car experiences while braking, and it’s the same for both cars.
Another way we can write this equation is as negative one over two times 𝑎 all multiplied by 𝑣 sub i squared. Written this second way, we can see that the braking distance 𝑠 is proportional to the initial speed of a vehicle squared. Mathematically, we can write that like this. 𝑠 is proportional to 𝑣 sub i squared.
Let’s now see how this proportionality affects the braking distances of our two vehicles. The braking distance of vehicle one, we’ll call that 𝑠 one, is equal to negative one over two times 𝑎 multiplied by 𝑣 sub one squared. We can recall that 𝑣 sub one is the initial speed of car one. Just as a side note, this negative sign we see appearing in the numerator is not a problem as far as our answer goes. Physically, it just means that the direction of our initial speed, in this case 𝑣 one, is opposite the direction of our acceleration 𝑎.
In comparing the braking distances of these two vehicles, what we’re really interested in is the quantity of the initial speed squared. Writing a similar equation for the braking distance of vehicle two, we’ll call that 𝑠 two, we see that it’s equal to negative one over two times 𝑎 times 𝑣 sub two squared. At this point, we can recall an important fact. The initial speed of vehicle one is twice that of vehicle two. That means in our equation for 𝑠 one, we can replace 𝑣 one with two times 𝑣 two. The square of two times 𝑣 two equals four times 𝑣 two squared.
Our equations for 𝑠 one and 𝑠 two are now written in terms of the same variables. If we divide 𝑠 one by 𝑠 two, we get a result of four. In other words, the braking distance of vehicle one is four times that of vehicle two. This result corresponds to answer option (B). That option says that driver one will have four times the braking distance of driver two because braking distance is proportional to the square of initial velocity.
Let’s look now at part two of our question.
How, if at all, will the thinking distances of the two vehicles be different? (A) They will be the same because thinking distance doesn’t depend on initial velocity. (B) Driver one will have half the thinking distance of driver two because thinking distance is inversely proportional to initial velocity. (C) Driver one will have four times the thinking distance of driver two because thinking distance is proportional to the square of initial velocity. (D) Driver one will have twice the thinking distance of driver two because thinking distance is proportional to initial velocity.
Earlier, we considered the braking distances of these two vehicles. And now we’re considering their thinking distances. The thinking distance is the distance traveled by the vehicles in the time it takes their drivers to react. We can define the reaction time of each driver as the difference in time from when light from the obstacle on the road first reaches the driver’s eyes to the instant that driver begins to press down on the brake pedal. Over that reaction time, a driver’s vehicle travels what is called the thinking distance.
Let’s assume that both of our drivers, like both of our vehicles, are identical in the sense that they have the same reaction time. That is, the amount of time it takes driver one to press down on the brake pedal after receiving light from the obstacle on the road is the same as that amount of time for driver two. Under this assumption, the thinking distances of our two vehicles would be the same if they were moving at the same initial speed. We can see this by recalling the relationship that speed is equal to distance divided by time. Rearranging this equation, we see that distance is equal to speed multiplied by time.
If we consider 𝑑 in this equation to be the thinking distance of a given vehicle, that distance is equal to the initial speed of the car 𝑣 multiplied by the reaction time of the driver in the car 𝑡. We’ve said that we’re assuming 𝑡 is the same for both drivers. They have the same reaction time. Therefore, if 𝑣, the initial speed of each vehicle, was the same, then 𝑑 would be the same in both cases.
However, we know that the initial speed of car one is actually twice that of car two. If we call 𝑑 two the thinking distance of vehicle two, then that’s equal to the initial speed of car two, that’s 𝑣 two, multiplied by the reaction time of the driver. We’ll call that 𝑡 sub r. We want to compare this to the thinking distance of vehicle one. We’ll call that 𝑑 one. This is equal to 𝑣 one multiplied by the same reaction time 𝑡 sub r. But then because 𝑣 one is equal to two times 𝑣 two, 𝑑 one is equal to two 𝑣 two times 𝑡 sub r. We see then that 𝑑 one, the thinking distance of vehicle one, is equal to twice the thinking distance of vehicle two.
Reviewing our answer options in light of this result, we see that it’s option (D), which says that driver one will have twice the thinking distance of driver two. That agrees with what we found here. But notice that there’s a second part of this statement. It says that driver one will have twice the thinking distance of driver two because thinking distance is proportional to initial velocity. We can see that this second part of answer option (D) is correct based on our equations for the thinking distances 𝑑 two and 𝑑 one. Both of these were equal to some value multiplied by the speed of that given vehicle to the first power.
Recall that these two equations came from the more general equation that 𝑑 is equal to 𝑣 times 𝑡. From this general equation, we can say that distance is directly proportional to speed. This confirms that indeed thinking distance is proportional to initial velocity. Our final choice then will be answer option (D). Driver one will have twice the thinking distance of driver two because thinking distance is proportional to initial velocity.
