# Calculating Stopping Distance and Reaction Time

When an object gains kinetic energy and changes position, it is known to be in motion. There are three types of motion possible, one dimensional, two-dimensional, and three-dimensional motion. One dimensional motion is the object moving in a straight line, two-dimensional motion is when an object moves covering two axes (x-y, y-z, z-x axis), three-dimensional motion is when an object moves in all three x-y-z directions. The motion of a body is observed w.r.t a frame of reference, on a graph, the origin is denoted as the frame of reference. Let’s learn about the Motion in a straight line in more detail,

### Motion in a Straight Line

Motion in one direction or motion in 1-dimension is the motion in a straight line. This type of motion is also known as Rectilinear motion. When the object has no kinetic energy and no external force is applied on the object, it remains at rest. When some external force is applied to the object and the object gains motion, it starts to move, if the motion occurs in one direction, it is known as rectilinear motion.

**Uniform Motion**

When the object travels with the same velocity throughout its motion, the motion is uniform. The initial and final speeds are the same in this case and the speed is given as,

**Speed = Distance/Time**

**Non-Uniform Motion**

When the object does not have the same velocity throughout its motion, it may increase its velocity or decrease it, such type of motion is known as Non-uniform motion. If the velocity keeps on decreasing, the object is decelerating, and if the velocity keeps on increasing, the object is accelerating. The formulae for non-uniform motion are given by Newton’s equation of motion.

First equation of motion ⇢ **v = u + at**

Second Equation of motion ⇢ **S = ut + 1/2(at ^{2})**

Third Equation of motion ⇢ ** v ^{2} = u^{2} + 2as**

Some very interesting concepts to see in Non-uniform motion are **Stopping distance and Reaction time. **Let’s learn about these two topics in more detail,

### Reaction Time

Reaction time can be defined as the time taken by a human to react or respond to a sudden change or stimulus. When people drive/ride their vehicle on the road, and suddenly an object comes in front of the road, now in order to stop the vehicle, the brain first has to acknowledge the hurdle on the way and warn the body to respond to the danger, the time taken to do this process is nothing but Reaction time.

Not only on road, normally reaction time can be analyzed too. Imagine holding a pen in hand at a certain height and asking a friend to instantly catch the pen as soon as it is dropped, it is obvious that the pen will fall a little before the friend can catch it. It is due to the fact that the friend is not aware of the time of the drop of a pen, hence, when the pen is dropped, he may take some time to respond to that, this is nothing but reaction time taken by the friend.

Reaction time depends on,

- The person taking the decision.
- The situations/surroundings in which the reaction time is taking place.

### Stopping Distance

When an object moves with a certain velocity and breaks are applied, the object takes some time to come to a stop, it is obvious that at this time, the object must have covered a certain distance, this distance is known as the Stopping distance. Imagine the same scenario as before, people driving/riding a vehicle and an object coming in front, after analyzing that brakes are needed to be applied, when the brakes are actually applied, it takes a certain time for the vehicle to come to rest, the distance covered in that time is Stopping Distance.

Stopping distance depends on:

- The velocity of the vehicle.
- The stopping capacity/braking capacity of the vehicle, that is, how fast the vehicle can decelerate.

**Deriving Formula for Stopping distance**

At the stopping point, the vehicle has applied the brakes, and now it has to come to a stop. The final velocity will become zero, hence, v = 0 is taken. Now, applying the third equation of motion in this case of non-uniform motion,

v^{2 }= u^{2 }+ 2aS

0 = u^{2 }+ 2aS

**S = -u ^{2}/2a**

Now, it is important to notice that a is taken in negative since the vehicle is decelerating. Hence, the stopping distance will finally come out to be positive.

**Deriving formula for Reaction time**

Imagine the case of a free-falling object and the time to react and catch it in order to find the reaction time. Keeping in mind that the initial velocity will be zero since the object is freely falling. Applying the second equation of motion in order to find the Reaction time of the vehicle,

S = ut + 1/2 (gt^{2})

u = 0m/sec

S = 1/2 (9.8 × t^{2})

t^{2 }= S × 0.204

**t ^{2 }= 0.204S**

**Sample Problems**

**Question 1: A car has a velocity of 20m/sec and the braking capacity of the car is given as 10m/sec ^{2}. What will be the Stopping distance for the car?**

**Solution:**

Since the acceleration occurs in negative direction, a = -10m/sec

^{2}Applying formula for stopping distance,

S = -u^{2}/2aS = -(20)

^{2}/2 × (-10)S = 400/20

S = 20 meters

**Question 2: How will the stopping distance be affected if the velocity of the vehicle is doubled?**

**Solution:**

Imagine in the first case, the velocity of the vehicle was v m/sec and the acceleration was a m/sec

^{2}.Now, stopping distance in the first case is given as,

S

_{1 }= -v^{2}/ 2aIn the second case, when the velocity is doubled, v’= 2v m/sec, acceleration = a m/sec

^{2}.Stopping distance is given as,

S

_{2 }= – (2v)^{2}/2aS

_{2 }= -(4v)/ 2aS

_{2 }= 4S_{1}Hence, if the velocity is doubled, the stopping distance increases 4 times.

**Question 3: A vehicle has a velocity of 25.5m/sec and the braking capacity of the vehicle is given as 16.5m/sec ^{2}. What will be the Stopping distance for the car?**

**Solution:**

Since the acceleration occurs in negative direction, a = -16.5m/sec

^{2}Applying formula for stopping distance,

S = -u^{2}/2aS = -(25.5)

^{2}/2 × (-16.5)S = 650.25/33

S = 19.704 meters

**Question 4: A car has a velocity of 19m/sec and the braking capacity of the car is given as 11m/sec ^{2}. What will be the Stopping distance for the car? **

**Solution:**

Since the acceleration occurs in negative direction, a = -11m/sec

^{2}Applying formula for stopping distance,

S = -u^{2}/2aS = -(19)

^{2}/2 × (-11)S = 361/22

S = 16.409 meters

**Question 5: An object is dropped and is now free-falling under gravity. The Stopping distance is given as 0.5 meters for the object. Find out the reaction time neglecting other factors.**

**Solution:**

The stopping distance, S = 0.5 m

The reaction time is given as,

t

^{2 }= 0.204St

^{2 }= 0.204 × 0.5t

^{2 }= 0.102t = 0.319 sec