Dabour Hub EGEgyptian Curriculum 2026/2027
2-1 (c) Acceleration as a function of position

2-1 (c) Acceleration as a function of position

Point 3 of 7 — Motion with Variable Acceleration and ForceQuestion bank: 0 questions (0 with pictures)

03 2-1 (c) Acceleration as a function of position

Method

  • If the force depends on the distance r from a fixed point, use m vdvdr=F(r).
  • Separate: ∫v0vv dv=1m∫r0rF dr, i.e. 12(v2−v02)=1m∫F dr.
  • To find the time afterwards, use v=drdt and integrate drv.
✍️ Worked examples

Example 6:

  • A body of mass 2 kg starts from rest at the origin under the force F=(8−2x) N along the x-axis (x in metres).
  • Find v in terms of x, the greatest speed, and where the body first stops.

Solution:

  • a=8−2x2=4−x, so vdvdx=4−x.
  • ∫0vv dv=∫0x(4−x) dx gives v22=4x−x22, so v2=8x−x2.
  • The speed is greatest where a=0, i.e. x=4 m: v2=32−16=16, so vmax=4 m/s.
  • The body stops when v=0: x(8−x)=0, so x=8 m. There the force is −8 N, so it turns back.

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