Showing posts with label Vehicle Motion. Show all posts
Showing posts with label Vehicle Motion. Show all posts

Sunday, July 12, 2026

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


 - Dr Sanjay Kumar Pawar

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

Educational infographic showing stopping distance formula with reaction distance, braking distance, motion equations, and a solved physics example for beginners.
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 Notes

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

Stopping Distance, Reaction Distance & Braking Distance Explained | NEET Physics Notes

- Dr Sanjay Kumar Pawar 

Reaction Distance, Braking Distance and Stopping Distance for NEET 

Educational diagram showing stopping distance, reaction distance, braking distance, reaction time, braking time, and stopping distance formula for NEET Physics.
Stopping distance consists of reaction distance and braking distance, an important concept in NEET Physics.


Internal Links

  1. Motion in One Dimension Notes
  2. Distance, Displacement and Speed
  3. Velocity and Acceleration Explained
  4. Equations of Motion (SUVAT Equations)
  5. Uniform and Non-uniform Motion
  6. Graphs of Motion (Distance-Time & Velocity-Time)
  7. Relative Motion Notes
  8. Newton's Laws of Motion
  9. Friction and Braking Force
  10. Work, Energy and Power
  11. Circular Motion
  12. NEET Physics Formula Sheet
  13. Top 100 NEET Physics MCQs
  14. Kinematics Practice Questions
  15. NEET Physics Revision Notes 
Stopping Distance Notes - NEET Physics

Stopping Distance (Reaction Distance + Braking Distance)

Definition

Stopping distance is the total distance travelled by a vehicle from the moment the driver notices an obstacle until the vehicle comes to complete rest.

Stopping Distance = Reaction Distance + Braking Distance
S = sr + sb

1. Reaction Distance (Thinking Distance)

Reaction distance is the distance travelled by the vehicle during the driver's reaction time before applying the brakes.

Reaction Time

Reaction time is the time taken by the driver from observing an obstacle to applying the brakes.

Generally, Reaction Time < 1 second

Formula

sr = u × tr

Where,

  • sr = Reaction Distance
  • u = Initial Velocity
  • tr = Reaction Time
During reaction time:
  • Brakes are not applied.
  • Velocity remains constant.
  • Acceleration = 0

2. Braking Distance

Braking distance is the distance travelled after applying the brakes until the vehicle comes to rest.

During braking,
  • Vehicle decelerates.
  • Final velocity becomes zero.

Formula

v² = u² − 2as
Since
v = 0
Therefore,
sb = u² / 2a
Where
  • sb = Braking Distance
  • a = Deceleration

3. Stopping Distance Formula

S = utr + u² / 2a
This is the most important formula for NEET examinations.

4. Safe Distance

Safe distance is the minimum distance required between two vehicles to avoid collision.

Safe Distance ≥ Stopping Distance

5. Difference Between Reaction Distance and Braking Distance

Reaction Distance Braking Distance
Before applying brakes After applying brakes
Velocity remains constant Velocity decreases
Acceleration = 0 Negative acceleration
Depends on reaction time Depends on braking force

6. Important NEET Points

  • Reaction distance depends on speed and reaction time.
  • Braking distance depends on speed².
  • If speed doubles, braking distance becomes four times.
  • Stopping distance = Reaction distance + Braking distance.
  • Reaction time is usually less than 1 second.

7. Formula Sheet

Reaction Distance = u × tr
Braking Distance = u² / 2a
Stopping Distance = utr + u² / 2a

8. Practice Questions

Q1. A car moves with speed 20 m/s. Reaction time is 0.5 s. Find the reaction distance.

Answer: 20 × 0.5 = 10 m
Q2. A vehicle moves at 30 m/s. Braking deceleration is 5 m/s². Find braking distance.
Answer: 30² / (2 × 5) = 900 /10 = 90 m
Q3. A car travels at 25 m/s. Reaction time = 0.8 s Deceleration = 5 m/s² Find stopping distance.
Reaction Distance = 25 × 0.8 = 20 m
Braking Distance = 25² /10 = 62.5 m
Stopping Distance = 20 + 62.5
= 82.5 m

9. Multiple Choice Questions

  1. During reaction time acceleration is
    • A. Positive
    • B. Negative
    • C. Zero ✔
    • D. Infinite
  2. Reaction distance depends upon
    • A. Speed
    • B. Reaction time
    • C. Both A and B ✔
    • D. Braking force
  3. Braking distance is proportional to
    • A. Speed
    • B. Speed² ✔
    • C. Time
    • D. Mass only
  4. Stopping distance equals
    • A. Reaction distance
    • B. Braking distance
    • C. Reaction distance + Braking distance ✔
    • D. None
  5. If speed doubles, braking distance becomes
    • A. Double
    • B. Half
    • C. Four Times ✔
    • D. Eight Times

10. One Minute Revision

✔ Reaction Distance = Before braking
✔ Braking Distance = After braking
✔ Stopping Distance = Reaction Distance + Braking Distance
✔ Reaction Distance = u × tr
✔ Braking Distance = u² / 2a
✔ Stopping Distance = utr + u² / 2a
✔ Braking Distance ∝ Speed²

Uniformly Accelerated Motion Class 11 Physics Notes | NEET & JEE MCQs

 - Dr.Sanjaykumar Pawar   Uniformly Accelerated Motion (1-D) Physics Notes, Formulas & NEET Questions  Uniformly Accelerated Motion (1-D...