Point 12 of 12 — Differentiation and Integration of Vector Functions and Newton’s Laws of MotionQuestion bank: 0 questions (0 with pictures)
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s = t³ − 3t² + 4 (m). v at t = 2 =0 m/s; a at t = 2 =6 m/s²
v = 2s² + 3. a = v dv/ds =(2s² + 3)(4s), so at s = 1: a = 20 m/s²
m = 3t + 2 kg, s = t²/2 m. v = t, so p = m v = 3t² + 2t; at t = 2, p =16 kg·m/sand F = dp/dt = 6t + 2 =14 N
r = (2t² − t) i + (t³) j. v at t = 1 =3 i + 3 j, |v| =3√2
A 4 kg body has v = 6 m/s and a force F = 2t N acts for 5 s. Δp = 25 kg·m/s, final velocity =12.25 m/s
3 kg-wt expressed in newtons =29.4 N; in dynes =2.94 × 10⁶ dyne
Uniform motion in a straight line implies the resultant force = zero
Distance from t = 0 to t = 3 when v = t² − 4t + 3: v vanishes at t = 1 and t = 3; ∫v = t³/3 − 2t² + 3t gives +4/3 on [0, 1] and −4/3 on [1, 3], so the distance is8/3 mwhile the displacement is0
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