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

2-1 (b) Acceleration as a function of time

Point 2 of 7 — Motion with Variable Acceleration and ForceQuestion bank: 0 questions (0 with pictures)

02 2-1 (b) Acceleration as a function of time

Method

  • If a=f(t), then v=v0+∫0tf(u) du and s=s0+∫0tv dt.
✍️ Worked examples

Example 1:

  • A particle has acceleration a=(6t−4) m/s2 and starts from the origin with velocity 2 m/s.
  • Find v and s at time t and their values at t=3.
  • Does the particle ever stop?

Solution:

  • v=2+∫0t(6u−4) du=3t2−4t+2.
  • s=∫0t(3u2−4u+2) du=t3−2t2+2t.
  • At t=3: v=27−12+2=17 m/s and s=27−18+6=15 m.
  • v=0 needs 3t2−4t+2=0, with discriminant 16−24<0: there is no solution, so the particle never stops.

Change of momentum

  • The change of momentum in [t1,t2] is ΔH=m∫t1t2a dt.
✍️ Worked examples

Example 4: A car of mass 1.5 tonnes has acceleration a=12t−t2 m/s2. Find the change of momentum in the first 6 s and in the interval [2,14].

Solution:

  • m=1500 kg. An antiderivative of a is 6t2−t33.
  • First 6 s: 6(36)−2163=216−72=144, so ΔH=1500×144=216 000 kg·m/s.
  • Interval [2,14]: at 14: 6(196)−27443=1176−914.67=261.33; at 2: 24−83=21.33.
  • Difference =240, so ΔH=1500×240=360 000 kg·m/s.

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