For constant acceleration, average velocity is:
vavg = u + v2
Displacement is:
s = vavg t
Therefore:
s = (u + v2) t
From the first equation:
v = u + at
Substituting v = u + at:
s = (u + (u + at)) / 2 × t
s = 2u + at2 × t
s = (u + ½at) t
s = ut + ½at²
Therefore:
s = ut + ½at²
13. Graphical Meaning of the Second Equation
Displacement equals the area under a velocity–time graph.
The rectangular area is:
Rectangle = ut
The triangular area is:
Triangle = ½(v − u)t
From the first equation:
v − u = at
Therefore:
Triangle = ½at²
Total displacement is:
s = ut + ½at²
14. Solved Example 2: Finding Displacement
Solved Example 2: Finding Displacement
A motorcycle starts from rest and accelerates at 3 m s⁻² for 4 s. Calculate its displacement.
1
Given
The motorcycle starts from rest: u = 0
a = 3 m s⁻²
t = 4 s
Required: s = ?
Formula: s = ut + ½at²
2
Working
s = (0)(4) + ½(3)(4)²
3
Step 2
s = 0 + ½(3)(16)
4
Step 3
s = 24 m
Final answer:s = 24 m
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Practice
Q1.An object has u = 5 m s⁻¹, a = 2 m s⁻², and t = 4 s. Find its displacement.
