Dabour Hub EGEgyptian Curriculum 2026/2027
Acceleration

Acceleration

Point 5 of 12 — Differentiation and Integration of Vector Functions and Newton’s Laws of MotionQuestion bank: 0 questions (0 with pictures)

05 Acceleration

Average and instantaneous

  • aav = Δv/Δt; the instantaneous acceleration is a = 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 is accelerated when |v| increases — that is when v and a have the same sign.
  • Motion is decelerated (retarded) when |v| decreases — when v and a have opposite signs.
  • a may be zero at an instant while the body is still moving: that happens at a maximum or a minimum velocity.

Worked examples

  1. 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.
  2. 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.
  3. 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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