Lesson Explainer: Reaction Time | Nagwa Lesson Explainer: Reaction Time | Nagwa

Lesson Explainer: Reaction Time Physics

In this explainer, we will learn how to describe the effects of a person’s reaction time on the motion of objects that they interact with.

Reaction time is the time taken for a human to respond to an incoming sensory signal. A classic experiment to measure your own reaction time requires only two things: a friend and a ruler. To perform it, have your friend hold the ruler at the very top and you hold your fingers at the zero mark. Your friend then drops it, and you then catch it as fast as you can, as shown below.

The difference in distance between where your fingers initially were, at the zero mark, and where you catch it, the catch point, gives you the total difference traveled. A shorter distance traveled means a faster reaction time, but the total distance traveled can be affected by multiple things: how straight your friend was holding the ruler, how slippery the ruler is, and how exactly at zero your fingers initially were.

These factors will give small differences, and the ruler is likely light enough that you can catch it nearly instantly, but when dealing with heavier objects, it can be more difficult to stop instantly. Let’s look at a different example of reaction time, when traveling in a car.

This car is traveling to the right, toward a wall. In this example, and all subsequent ones, we are assuming that our moving objects are not changing direction, so we only care about the magnitude, speed. We will use distance and displacement interchangeably.

The driver of the car, upon realizing they are moving toward a wall, would like to stop and so begins braking. Since they are traveling at some speed, however, they will travel a certain distance forward before they are able to react. This distance is called the thinking distance.

The time between being where we first start measuring, π‘‘οŠ¦, and where the car begins braking, 𝑑brake, is called the reaction time.

As an equation, looking at the car’s speed, 𝑣, the distance 𝑑 it travels over a period of time 𝑑 is 𝑑=𝑣𝑑.

When applied to the scenario above, note that this equation only works as long as the speed 𝑣 is constant. This means it only applies during the reaction time, since the car has not begun to brake yet. The distance would thus be the thinking distance. So the equation would look more like 𝑑=𝑒𝑑,thinkingreaction where 𝑑thinking is the thinking distance, 𝑒 is the initial speed, and 𝑑reaction is the reaction time. We can see that, in this scenario, 𝑒 is directly proportional to 𝑑thinking: doubling speed means doubling the distance covered.

When the driver uses the brakes, they do not stop instantly. The quality of the brakes, slipperiness of the road, and type of tires the car has are all factors that can affect how quickly the car stops. The distance the car travels after braking is called the braking distance.

The time from when the car begins braking, 𝑑brake, to when it comes to a complete stop, 𝑑stop, is called the braking time. Unlike reaction time, the speed of the car 𝑣 is not constant, since it is continually decreasing, so we cannot use the simple 𝑑=𝑣𝑑 equation to express it.

Instead, for an object with a constant acceleration, we can express its current speed as 𝑣=𝑒+2π‘Žπ‘‘, where 𝑣 is the final speed, 𝑒 is the initial speed, π‘Ž is acceleration, and 𝑑 is the distance it travels.

If we assume that the car is decelerating at a constant rate, all the way down to a stop, we can adjust the equation as follows: 0=𝑒+2π‘Žπ‘‘,braking where 𝑒 is the initial speed, π‘Ž is the acceleration rate of the car, and 𝑑braking is the braking distance. The final speed, 𝑣, has been replaced with a 0 since the car’s final speed will be 0, when it stops. We can then subtract π‘’οŠ¨ from both sides, 0βˆ’π‘’=𝑒+2π‘Žπ‘‘βˆ’π‘’,braking to get βˆ’π‘’=2π‘Žπ‘‘.braking

Now we have a negative in front of the initial speed, but this is not a problem, since we know the value of acceleration is negative too. This makes the equation 𝑒=2|π‘Ž|𝑑.braking

This equation shows us that the 𝑑braking is proportional to the square of the initial speed 𝑒. Doubling the initial speed means quadrupling the braking distance. We see from this equation also that reaction time is in no way related to the braking distance.

The total distance that the car has taken to stop (the stopping distance) is affected by its initial speed, for both thinking distance and braking distance.

The total time taken to stop (the stopping time) is affected by the reaction time of the driver, and the initial speed and the deceleration rate of the brakes for the braking time.

Let’s look at an example.

Example 1: Braking Time Bars

The thinking distance and braking distance for a car at different initial speeds are shown by the lengths of the two-colored bars in the diagram. The longer a bar, the greater the initial speed the car stops from. Which of the following quantities is shown by the length of the gray part of the bar?

  1. Stopping distance
  2. Thinking distance
  3. Braking distance

Answer

The total distance it takes for the car to stop would be the stopping distance, which would be a combination of both the thinking and the braking distance. This means it would be both the purple and grey bars, not just one or the other.

Thinking distance is proportional to the initial speed of the car. Since, on these bars, greater initial speed is represented by greater total bar lengths, we should expect whichever bar is thinking distance to increase linearly.

Braking distance is proportional to the square of the initial speed of the car. On these bars, we should see the braking distance increase faster for higher initial velocities.

The purple part of the bars starts small but increases dramatically as the initial velocity increases. The grey part of the bars increases at about the same rate as initial velocity increases. This means that the grey part is not a square relation, making it the thinking distance.

The correct answer is b, thinking distance.

Due to how both the thinking distance and the braking distance increase with velocity, the greatest factor that affects the total stopping distance is just initial velocity. Of course, improving the reaction time and deceleration rate will decrease the stopping distance as well, but not moving at a high speed in the first place is far easier.

Consider the deceleration rate: the efficiency of the brakes, road conditions, and general car quality are all things that typically degrade or get worse, rather than get better, and they all increase the braking distance.

The difference in these conditions can mean a dramatic difference in braking distance, perhaps enough to not brake in time.

Let’s look at an example.

Example 2: Braking Car Velocity–Time Graph

The velocity–time graph shows the change in the velocity of a car that suddenly brakes to come to a stop on a dry concrete surface.

Which of the other graphs shown, (a), (b), (c), (d), and (e), best matches the velocity–time graph for the same car stopping, driven by the same driver, but where the car travels on a wet concrete surface?

Answer

Let’s first carefully examine the language of the question: we are looking for β€œthe velocity–time graph for the same car stopping, driven by the same driver, but where the car travels on a wet concrete surface.”

Same car means the brakes are the same, and same driver means the same reaction time. The only difference is the surface, with wet concrete taking longer to brake on than dry concrete. This would show up on a graph as a less steep line, as the deceleration rate would be lower. Graphs (d) and (e) seem to maintain a steep line, so they likely are not it.

Since the reaction time is the same, the initial distance where the velocity is constant (the thinking distance) should be the same across all graphs. Right away we see it is not: (a), (b), and (d) all have a longer thinking distance than the base graph, taking them right out.

This brings us down to graphs (c) and (e). As previously mentioned, (e) has a rather steep line, but looking closely, we also see that it has a higher initial velocity! So (e) cannot be it.

Graph (c) has the same reaction time and initial velocity, but has a longer braking distance, as evidenced by the more gradual incline.

The correct answer is thus (c).

Let’s begin looking at some cars slowing down using numbers.

Recall first the two equations we used earlier in this explainer for thinking distance, 𝑑thinking𝑑=𝑒𝑑,thinkingreaction and braking distance, 𝑑braking, 𝑒=2|π‘Ž|𝑑,braking where 𝑒 is the initial speed, 𝑑reaction is the reaction time, and π‘Ž is the rate of deceleration.

These equations can be used to determine the total stopping distance, as stopping distance is just thinking distance and braking distance added together: 𝑑=𝑑+𝑑.stoppingthinkingbraking

Now let’s determine the stopping distance of a car that is hurtling toward a cliff. Its initial velocity is 25 m/s, deceleration once brakes are applied is 4 m/s2, and the driver’s reaction time is 1.25 s.

We should start by individually finding the values of 𝑑thinking and 𝑑braking. Since 𝑑thinking is by itself, all we need to do is put in our known values: 𝑑=𝑒𝑑.thinkingreaction

Our initial velocity, 𝑒, is 25 m/s. The reaction time, 𝑑reaction, is 1.25 s. Putting these in yields 𝑑=(25/)(1.25).thinkingmss

The metre per second (m/s) unit in velocity become just metres when multiplied by seconds, so we have (25/)(1.25)=31.25.mssm

Thinking distance, 𝑑thinking, is thus 31.25 metres.

Now, to obtain the value of 𝑑braking, we first have to isolate it in its equation: 𝑒=2|π‘Ž|𝑑.braking

We do this by dividing both sides by 2|π‘Ž|, which should cancel the 2|π‘Ž| on the right side: 𝑒2|π‘Ž|=2|π‘Ž|𝑑2|π‘Ž|.braking

The 2|π‘Ž| factor on the right side divides and cancels, leaving us with 𝑒2|π‘Ž|=𝑑.braking

Since we know the values of the initial velocity 𝑒 as 25 m/s and deceleration π‘Ž as 4 m/s2, we just have to put those in the equation now: (25/)2(4/)=𝑑.msmsbraking

Let’s square the initial velocity and multiply the 2 with the deceleration ο€Ή625/(8/)=𝑑.msmsbraking

Remember that squaring a term also squares its units. This means we have units of square metres per second squaredο€Ή/ms on the top and metres per second squaredο€Ή/ms on the bottom. Dividing through, we obtain the following in units of just metres: ο€Ή625/(8/)=78.125.msmsm

Braking distance, 𝑑braking, is thus 78.125 metres.

Taking the sum of these together to get the stopping distance, we have 𝑑=31.25+78.12531.25+78.125=109.375.stoppingmmmmm

The stopping distance for this car is thus 109.375 metres.

Let’s look at another example.

Example 3: Braking Car at a Bridge

A car drives along a road at 15 m/s toward a bridge, as shown in the diagram. When the front wheels of the car are 50 m from the bridge, the driver sees a sign warning that the bridge has a collapsed section. The car can decelerate at 5 m/s2. What is the maximum reaction time that the driver can have and still stop the car before it reaches the bridge? Answer to one decimal place.

Answer

This time, we are not finding the stopping distance, we are already given it: 50 m/s. We are trying to find the maximum reaction time of the driver instead. Reaction time is in the equation that also contains initial velocity and braking distance: 𝑑=𝑒𝑑.thinkingreaction

Let’s isolate the reaction time by dividing both sides by the initial velocity 𝑒: 𝑑𝑒=𝑒𝑑𝑒.thinkingreaction

The 𝑒’s cancel on the right side, leaving just 𝑑𝑒=𝑑.thinkingreaction

We are given the initial velocity 𝑒, of 15 m/s, so we need to find the thinking distance. Luckily, we know that it is related to the stopping distance: 𝑑=𝑑+𝑑.stoppingthinkingbraking

We already know the stopping distance, 50 m, so we just need to determine braking distance. Braking distance can be determined by the other equation of 𝑒=2|π‘Ž|𝑑.braking

To isolate the braking distance, we divide both sides by 2|π‘Ž|, which should cancel the 2|π‘Ž| on the right side: 𝑒2|π‘Ž|=2|π‘Ž|𝑑2|π‘Ž|.braking

2|π‘Ž| cancels, so we have 𝑒2|π‘Ž|=𝑑.braking

We know the initial speed, 𝑒, is 15 m/s and the rate of deceleration, π‘Ž, is 5 m/s2. Putting in these values gives (15/)2(5/)=𝑑.msmsbraking

Squaring through and multiplying the 2 by the value of deceleration gives ο€Ή225/10/=𝑑.msmsbraking

The units cancel to become just metres, leaving us with ο€Ή225/10/=22.5.msmsmetres

The braking distance is thus 22.5 metres. Putting this into the equation for the total stopping distance of 50 m, we see we have 𝑑=𝑑+𝑑50=𝑑+22.5.stoppingthinkingbrakingthinkingmm

Subtracting both sides by 22.5 metres gives us the value of thinking distance: 50βˆ’22.5=𝑑50βˆ’22.5=27.5.mmmmmthinking

Now that we have our value of thinking distance, we can finally take it back to our initial equation of reaction time: 𝑑𝑒=𝑑.thinkingreaction

Putting in the values of 27.5 metres for thinking distance and 15 m/s for initial speed gives 27.515/=𝑑.mmsreaction

The units of metres divided by metres per second give just seconds, producing our maximum reaction time of 27.515/=1.8333333….mmss

To one decimal place, the maximum time the driver has to react is 1.8 seconds.

There is another way we can understand the stopping of a car besides the total distance traveled, and that is with the total time it takes to stop.

To see what this is about, let’s look back at the example of the car hurtling toward the cliff, but instead look a the time.

We know that the total time taken to stop (stopping time) is 𝑑=𝑑+𝑑.stoppingreactionbraking

We are given the reaction time, 𝑑reaction, of 1.25 seconds, but the time taken for the car to brake from its current speed to a complete stop is not given.

The braking time is found by using a different kinematic equation that takes advantage of constant acceleration. A constant deceleration from an initial speed can be related to time as follows: 𝑒=|π‘Ž|𝑑,braking where 𝑒 is the initial velocity of the car and π‘Ž is the deceleration of the car, once again using the absolute value. We still need to isolate the value of 𝑑braking though, so let’s do this by dividing both sides by |π‘Ž|: 𝑒|π‘Ž|=|π‘Ž|𝑑|π‘Ž|.braking

The |π‘Ž|’s on the right side cancel, giving 𝑒|π‘Ž|=𝑑.braking

Since we know the initial speed, 25 m/s, and deceleration, 4 m/s2, we just need to put these into the equation to obtain our value of braking time: 25/4/=𝑑.msmsbraking

The units of metres per second(/)ms divided by metres per second squaredο€Ή/ms give back seconds, so the time is 25/4/=6.25.msmss

The car takes a harrowing 6.25 seconds to slow down to a complete stop when traveling at 25 m/s. Combining this with the reaction time gives us the total time taken: 𝑑=6.25+1.25𝑑=7.5.stoppingstoppingsss

The stopping time is a full 7.5 seconds for this car.

Let’s look at another example.

Example 4: Braking Car Reaction Time

The driver of a car traveling at 30 m/s has a reaction time of 1.5 s. The car’s brakes decelerate the car at 3.75 m/s2 once they are activated. How much time does the car take to stop, including the driver’s thinking time?

Answer

The total time it takes for the car to stop is the sum of the reaction time and braking time: 𝑑=𝑑+𝑑.stoppingreactionbraking

We are given the reaction time of 1.5 seconds, but we still need to find the braking time. To do this, we will use the equation that relates braking time to two values we are already given, the deceleration and initial speed: 𝑒=|π‘Ž|𝑑.braking

Solving for braking time, we divide both sides by |π‘Ž|: 𝑒|π‘Ž|=|π‘Ž|𝑑|π‘Ž|,braking which simplifies to 𝑒|π‘Ž|=𝑑.braking

Let’s put in our values of initial speed, 30 m/s, and deceleration, 3.75 m/s2: 30/3.75/=𝑑.msmsbraking

The units of metres per second (m/s) divided by metres per second squaredο€Ή/ms become just seconds, making the braking time 30/3.75/=8.msmss

Together with the reaction time, the total time it takes for the car to stop is thus 𝑑=1.5+8𝑑=9.5.stoppingstoppingsss

The total time spent stopping is 9.5 seconds.

Now that we have seen how to find the total distance and total time it takes for a car to stop, let’s see how these values would look when represented graphically.

Let’s look at some examples.

Example 5: Rate of Velocity Change

The graph shows how the stopping distances of two cars change depending on the speeds at which the cars are moving at the moment they start to decelerate.

  1. What is the difference in the stopping distances of the cars when they both stop at the speed of 15 m/s?
  2. What is the difference in the stopping distances of the cars when they both stop at the speed of 20 m/s?

Answer

Part 1

To find the difference in the stopping distances of the cars when they stop at a speed of 15 m/s, we must first find their stopping distances individually. The car represented by the blue line has a stopping distance of about 30 m with a speed of 15 m/s, and the orange line car has a stopping distance of about 12 m. The difference between these two is just 30βˆ’12=18.mmm

So, the difference in stopping distance is 18 metres when the cars are initially traveling at 15 m/s.

Part 2

When at 20 m/s, the car represented by the blue line has a stopping distance of 50 metres. The car represented by the orange line has a stopping distance of 20 metres. The difference is thus 50βˆ’20=30.mmm

At 20 m/s, the stopping distance differs by 30 metres.

Compared with the first difference in stopping distance, the difference is much greater. Traveling at higher speeds means greater stopping distances.

Example 6: Thinking and Braking Distances for a Car

The thinking distance and braking distance for a car at different initial speeds are shown in the graph. The thinking distance is in blue and the braking distance is in orange.

  1. What is the lowest speed, to the nearest kilometre per hour, where the braking distance is greater than the thinking distance?
  2. What is the stopping distance for an initial speed of 50 km/h?
  3. What is the stopping distance for an initial speed of 80 km/h?
  4. How much greater is the braking distance than the thinking distance when the initial speed is 90 km/h?
  5. How much lower is the braking distance than the thinking distance when the initial speed is 40 km/h?

Answer

Part 1

The lowest speed where the braking distance, the orange line, is greater than the thinking distance, the blue line, is around an initial speed of 75 km/h. This is where the lines meet, and the braking distance eclipses the thinking distance.

Part 2

The stopping distance for an initial speed of 50 km/h is equal to the sum of the thinking and braking distances. At 50 km/h, the thinking distance is around 14 metres, and the braking distance is 21 metres. The stopping distance is these added together, which is 35 metres.

Part 3

Looking at the graph for an initial speed of 80 km/h, the thinking distance is 33 metres, and the braking distance is 36 metres. The sum of these, the stopping distance, is thus 69 metres.

As initial speed increases, the total distance contribution for both thinking and braking distances increases, but braking distance increases faster.

Part 4

At an initial speed of 90 km/h, the braking distance is 45 metres. The thinking distance is 37 metres, so the difference between the two is 8 metres. The braking distance is 8 metres greater at 90 km/h.

Part 5

When the initial speed is 40 km/h, braking distance is 9 metres, and the thinking distance is 17 metres. The difference between the two is 8 metres, so the braking distance is lower than the thinking distance by 8 metres.

Again, braking distance, being a squared relationship, increases faster than thinking distance as initial speed increases.

Let’s summarize what we have learned in this explainer.

Key Points

  • Reaction time represents the quickness of human response to an incoming sensory signal.
  • For a vehicle, there is a distance taken to begin stopping, the thinking distance, and a distance during which deceleration occurs, braking distance.
  • The thinking distance is proportional to the initial speed of the vehicle.
  • The braking distance is proportional to the square of the initial speed of the vehicle.
  • Only thinking distance is proportional to reaction time.

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