Stopping Distance Formula
Last Updated : 09 Apr, 2025
Have you ever observed that a moving vehicle does not come to rest as soon as its brakes are applied after it has been moving at a constant or even varying velocity for quite some time?
When a motorist applies the brakes to an automobile, it does not come to a complete stop. The stopping represents the distance the automobile travels before coming to a complete stop. It is determined by the vehicle's speed and the friction coefficient between the tyres and the road. This calculation does not account for the effect of pro brakes or brake pumps.
The stopping represents the distance travelled between the moment the person wants to halt a moving vehicle and the time the vehicle comes to a complete stop. The stopping distance is represented by d and is affected by factors such as road pavement and the driver's responses. The stopping distance is measured in meters.
Formula of Stopping Distance,
d=\frac{v^2}{2\mu g}
where,
- 'v 'denotes the velocity in ms-1
- 'µ' denotes the friction coefficient
- 'g ' is the acceleration owing to gravity
OR
d = kv2
where,
- g = acceleration due to gravity ( 9.8 d =kv2 )
- k is the proportionality constant
- v refers to the velocity of the vehicle
Solved Examples
Problem 1. Find the stopping distance of a truck given its speed of 70 km/hr. and coefficient of friction 0.9.
Solution:
Given: µ = 0.9 and v = 70 km/hr. or 19.44 ms-1
Since, d=\frac{v^2}{2\mu g}\\= \frac{(19.44)^2}{2(0.9)(9.8)}
= 377.9136/(1.8 × 98)
d = 21.42 m
Problem 2. Find the stopping distance of a motorbike given its velocity is 13 ms-1 and the proportionality constant is 0.5.
Solution:
Given: v = 13 ms-1 and k = 0.5
Since, d = kv2
= (0.5)(13)2
= 0. 5 × 169
d = 84.5 m
Problem 3. Find the proportionality constant if a car travelling at 30 m/s covers a distance of 20 m before coming to rest when brakes are applied.
Solution:
Given: v = 30 m/s and d = 20 m
Since, d = kv2
⇒ k = d/v2
= 20/(30)2
= 20/900
k = 0.0222
Problem 4. Find the stopping distance of a motorbike given its velocity is 10 ms-1 and the proportionality constant is 0.5.
Solution:
Given: v = 10 ms-1 and k = 0.5
Since, d = kv2
= (0.5)(10)2
= 0. 5 × 100
d = 50 m
Problem 5. Find the stopping distance of a motorbike given its velocity is 24 ms-1 and the proportionality constant is 0.5.
Solution:
Given: v = 24 ms-1 and k = 0.5
Since, d = kv2
= (0.5)(24)2
= 0. 5 × 576
d = 288 m
Conclusion
The formula for stopping distance is d= v2/2μg , where 'd' is the stopping distance, 'v' is the initial velocity,' μ' is the coefficient of friction, and 'g' is the acceleration due to gravity. This equation highlights the importance of road surface conditions and vehicle speed in determining how quickly a vehicle can stop.
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