Point 5 of 12 — Differentiation and Integration of Vector Functions and Newton’s Laws of MotionQuestion bank: 0 questions (0 with pictures)
05Acceleration
Average and instantaneous
aav = Δv/Δt; the instantaneous acceleration isa = dv/dt = d²s/dt², the slope of the tangent to the (velocity–time) curve. Unit: m/s² (or cm/s²).
From s(t): differentiate once for v, twice for a.
Acceleration as a function of position
When v is given in terms of s, use the chain rule: a = dv/dt = (dv/ds)(ds/dt) = v dv/ds, equivalently a = d/ds (½v²).
Accelerated and decelerated motion
Motion isacceleratedwhen |v| increases — that is when v and a have the samesign.
Motion isdecelerated (retarded)when |v| decreases — when v and a haveoppositesigns.
a may be zero at an instant while the body is still moving: that happens at a maximum or a minimum velocity.
Worked examples
A particle moves on a line with s = t³ − 6t² + 9t (metres). v = 3t² − 12t + 9 = 3(t − 1)(t − 3), a = 6t − 12. (a) v = 0 at t = 1 and t = 3: a(1) =−6 m/s², a(3) =+6 m/s². (b) a = 0 at t = 2: v = 12 − 24 + 9 =−3 m/s, so the speed is 3 m/s.
(c) Direction changes at t = 1 and t = 3 with s(0) = 0, s(1) = 4, s(3) = 0, s(4) = 4, so distance = 4 + 4 + 4 = 12 m.
A particle moves with v = 2s + 1. Find a when s = 3 m: a = v dv/ds = (2s + 1)(2), so at s = 3, a = 7 × 2 =14 m/s².
Same idea with a square root: if v = √(2s + 1), then ½v² = s + ½ and a = d/ds(½v²) =1 m/s², the same at every position.
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