Stopping Distance Formula Explained with Solved Examples | Reaction & Braking Distance

 - Dr Sanjay Kumar Pawar

Stopping Distance: Formula, Reaction Distance, Braking Distance & Questions

Stopping Distance consists of Reaction Distance and Braking Distance. Learn the formula with an easy solved example.

Internal Links

• Motion in a Straight Line
• Equations of Motion
• Uniform and Non-Uniform Motion
• Speed, Velocity and Acceleration
• Distance and Displacement
• Newton's Laws of Motion
• Friction in Physics
• Work, Energy and Power
• Projectile Motion
• Free Fall Equations
• Average Speed Formula
• Relative Motion
• SUVAT Equations
• Kinematics Notes
• Physics Formula Sheet

Stopping Distance (Reaction Distance + Braking Distance)

What is Stopping Distance?

When a driver sees an obstacle, the car does not stop immediately. The car travels some distance before stopping completely.

The stopping distance has two parts:

• Reaction Distance → Distance travelled while the driver reacts.
• Braking Distance → Distance travelled after applying the brakes.

Stopping Distance = Reaction Distance + Braking Distance

S = Sr + Sb

1. Reaction Distance

Sr = u × t

Where

• u = Initial speed (m/s)
• t = Reaction time (seconds)

2. Braking Distance

Use the equation of motion:

v² = u² + 2as

When the car stops:

• v = 0
• a = Retardation

Therefore,

Sb = u² / 2a

Solved Example

Question

A car is moving with velocity 20 m/s. The driver's reaction time is 0.5 s. The maximum retardation is 4 m/s². Find the stopping distance.

Step 1 : Reaction Distance

Sr = u × t

Sr = 20 × 0.5

Sr = 10 m

Step 2 : Braking Distance

Sb = u² / 2a

= 20² / (2 × 4)

= 400 / 8

= 50 m

Step 3 : Stopping Distance

S = Sr + Sb

= 10 + 50

= 60 m

Answer = 60 m

Shortcut Formula

Stopping Distance

S = ut + u² / 2a

Important Points

• Reaction distance depends on reaction time.
• Braking distance depends on speed and braking force.
• Higher speed means longer stopping distance.
• Wet roads increase braking distance.
• Always maintain a safe distance while driving.

Practice Questions

Q1. A car moves at 10 m/s. Reaction time = 1 s Retardation = 5 m/s² Find stopping distance.

Q2. A car moves at 30 m/s. Reaction time = 1 s Retardation = 6 m/s² Find stopping distance.

Q3. A car moves at 15 m/s. Reaction time = 0.4 s Retardation = 3 m/s² Find stopping distance.

Q4. A car moves at 25 m/s. Reaction time = 0.8 s Retardation = 5 m/s² Find stopping distance.

Answers

Question	Answer
Q1	20 m
Q2	105 m
Q3	43.5 m
Q4	82.5 m

Quick Revision

Quantity	Formula
Reaction Distance	Sr = ut
Braking Distance	Sb = u² / 2a
Stopping Distance	S = Sr + Sb

Memory Trick

Think → Brake → Stop

• Think = Reaction Distance
• Brake = Braking Distance
• Stop = Stopping Distance

© Physics Notes for Beginners

Physics Example 4.2 & 4.3 Solutions | Newton's Second Law Step by Step

 - Dr.Sanjaykumar pawar

Step-by-step solutions for NCERT Physics Examples 4.2 and 4.3 demonstrating Newton's Second Law and force calculations.

Physics Solved Examples

Example 4.2

Question:
A bullet of mass 0.04 kg moving with a speed of 90 m/s enters a heavy wooden block and is stopped after travelling a distance of 60 cm. What is the average resistive force exerted by the block on the bullet?

Given:

• Mass of bullet, m = 0.04 kg
• Initial velocity, u = 90 m/s
• Final velocity, v = 0 m/s (bullet stops)
• Distance travelled, s = 60 cm = 0.6 m

To Find:

Average resistive force exerted by the block on the bullet.

Step 1: Use the equation of motion
v² = u² + 2as

Substitute the values:
0² = 90² + 2(a)(0.6)
0 = 8100 + 1.2a
1.2a = -8100
a = -8100 / 1.2
a = -6750 m/s²

The negative sign indicates retardation (deceleration).

Step 2: Apply Newton's Second Law
F = ma

F = 0.04 × 6750
F = 270 N

Final Answer: Average resistive force = 270 N

Example 4.3

Question:
The motion of a particle of mass m is described by the equation: y = ut + ½gt². Find the force acting on the particle.

Given:

y = ut + ½gt²

To Find:

Force acting on the particle.

Step 1: Find Velocity
Velocity is the rate of change of displacement.

v = dy/dt

Differentiate y = ut + ½gt²
v = u + gt

Step 2: Find Acceleration
Acceleration is the rate of change of velocity.

a = dv/dt

Differentiate v = u + gt
a = g

Step 3: Apply Newton's Second Law
F = ma

Since a = g,
F = mg

Final Answer: F = mg

Therefore, the particle moves under the influence of gravitational force.