Topic 1 of 8 · Form 5
Kinematics
Velocity v = ds/dt; acceleration a = dv/dt = d²s/dt². Conversely, s = ∫v dt and v = ∫a dt (use initial conditions for constants).
Notes for this topic 1 file, in order
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Displacement, velocity and acceleration
Velocity v = ds/dt; acceleration a = dv/dt = d²s/dt². Conversely, s = ∫v dt and v = ∫a dt (use initial conditions for constants).
Constant acceleration (suvat)
v = u + at; s = ut + ½at²; v² = u² + 2as; s = ½(u + v)t; s = vt − ½at².
Worked example: A car decelerates uniformly from 25 m s⁻¹ to rest in 50 m. Find the deceleration and the time taken.
0 = 625 + 2a(50) → a = −6.25 m s⁻² (deceleration 6.25 m s⁻²)
0 = 25 − 6.25t → t = 4 s
Variable acceleration
Worked example: A particle moves with v = 3t² − 12t + 9 m s⁻¹. Find when it is at rest and its acceleration at t = 2.
3(t² − 4t + 3) = 0 → t = 1 s or t = 3 s
a = 6t − 12 → at t = 2, a = 0 m s⁻²
Graphs
Gradient of a displacement–time graph = velocity; gradient of a velocity–time graph = acceleration; area under a v–t graph = displacement.
Relative velocity (vectors)
vA relative to B = vA − vB; closest approach problems use the relative position vector.
Check yourself
- A ball is dropped from 45 m (g = 10 m s⁻²). Find the time to reach the ground.
- v = 4t − t². Find the displacement from t = 0 to t = 3.
- What does the area under a velocity–time graph represent?
Show answers
- 45 = 5t² → t = 3 s
- [2t² − t³/3] from 0 to 3 = 18 − 9 = 9 m
- The displacement.
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