Average velocity is:
vavg = u + v2
Displacement is:
s = (u + v2) t
From the first equation:
v = u + at
Rearranging for time:
v − u = at
t = v − ua
Substituting this value of t:
s = (u + v2) × (v − ua)
Using:
(u + v)(v − u) = v² − u²
Therefore:
s = v² − u²2a
Multiplying both sides by 2a:
2as = v² − u²
Adding u² to both sides:
v² = u² + 2as
Therefore:
v² = u² + 2as
17. Solved Example 3: Braking Distance
Solved Example 3: Braking Distance
A car moving at 20 m s⁻¹ comes to rest with a uniform acceleration of −5 m s⁻². Find its stopping distance.
1
Given
u = 20 m s⁻¹
The car comes to rest: v = 0
a = −5 m s⁻²
Required: s = ?
Formula: v² = u² + 2as
2
Working
(0)² = (20)² + 2(−5)s
3
Step 2
0 = 400 − 10s
4
Step 3
10s = 400
5
Step 4
s = 40 m
6
Step 5
Key Step: The acceleration is negative because it acts opposite to the car’s velocity.
Final answer:s = 40 m
Did you understand this topic?
Your answer is saved with your learning record. It does not mark the topic as mastered.
Ready to test yourself?
You've completed the learning section.
1 concept questions
Easy → Medium → Hard
Explanation after each answer
Practice
Practice
Q1. A vehicle slows from 25 m s⁻¹ to 5 m s⁻¹ at −4 m s⁻². Find the distance travelled.
Practice
Q2. Derive the third equation by eliminating time.
