Solutions — Displacement and Velocity Applications

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1. Show Solution
   (a) s(t) = ∫(4t − 8) dt = 2t² − 8t + c
   s(0) = 0 ⇒ c = 0
   s(t) = 2t² − 8t
   
   (b) v(t) = 0: 4t − 8 = 0 ⇒ t = 2 s
   
   (c) Displacement = s(6) − s(0) = (2 × 36 − 48) − 0 = 72 − 48 = 24 m
2. Show Solution
   (a) s(t) = ∫(3 − 6t) dt = 3t − 3t² + c
   s(0) = c = 5
   s(t) = 3t − 3t² + 5
   
   (b) s(3) = 9 − 27 + 5 = −13 m
   (The particle is 13 m in the negative direction from the origin.)
3. Show Solution
   Solve v(t) = 0: t² − 4 = 0 ⇒ t² = 4 ⇒ t = ±2
   Since t ≥ 0, the particle is at rest at t = 2 s.
   
   Check: v(1) = 1 − 4 = −3 < 0 (moving in negative direction before t = 2)
   v(3) = 9 − 4 = 5 > 0 (moving in positive direction after t = 2)
4. Show Solution
   v = 0 at t = 2 (from Q1). Sign diagram:
   v < 0 on [0, 2] (particle moving left)
   v > 0 on [2, 6] (particle moving right)
   
   Positions: s(0) = 0, s(2) = 8 − 16 = −8, s(6) = 72 − 48 = 24
   
   Total distance = |s(2) − s(0)| + |s(6) − s(2)|
   = |−8 − 0| + |24 − (−8)|
   = 8 + 32 = 40 m
5. Show Solution
   (a) v(t) = 0: t² − 4t + 3 = (t − 1)(t − 3) = 0
   At rest at t = 1 s and t = 3 s.
   
   (b) s(t) = ∫(t²−4t+3) dt = t³/3 − 2t² + 3t + c
   s(0) = c = 2
   s(0) = 2
   s(1) = 1/3 − 2 + 3 + 2 = 1/3 + 3 = 10/3
   s(3) = 27/3 − 18 + 9 + 2 = 9 − 18 + 11 = 2
   s(4) = 64/3 − 32 + 12 + 2 = 64/3 − 18 = 64/3 − 54/3 = 10/3
   
   Total distance = |10/3 − 2| + |2 − 10/3| + |10/3 − 2|
   = 4/3 + 4/3 + 4/3 = 4 m
6. Show Solution
   s(t) = ∫ 6e^−2t dt = 6 × (1/(−2)) e^−2t + c = −3e^−2t + c
   s(0) = −3e⁰ + c = −3 + c = 4 ⇒ c = 7
   s(t) = −3e^−2t + 7
   
   s(2) = −3e⁻⁴ + 7 = 7 − 3e⁻⁴ m ≈ 6.945 m
7. Show Solution
   (a) v(t) = ∫(6t − 6) dt = 3t² − 6t + c
   v(0) = 0 ⇒ c = 0. v(t) = 3t² − 6t = 3t(t − 2)
   
   (b) s(t) = ∫(3t² − 6t) dt = t³ − 3t² + c
   s(0) = 0 ⇒ c = 0. s(t) = t³ − 3t²
   
   (c) v = 0 at t = 0 and t = 2.
   s(0) = 0, s(2) = 8 − 12 = −4, s(3) = 27 − 27 = 0
   Total distance = |−4 − 0| + |0 − (−4)| = 4 + 4 = 8 m
8. Show Solution
   Find v(t):
   v(t) = ∫ 2 dt = 2t + c; v(0) = 5 ⇒ c = 5
   v(t) = 2t + 5
   
   Find s(t):
   s(t) = ∫(2t + 5) dt = t² + 5t + k; s(0) = −3 ⇒ k = −3
   s(t) = t² + 5t − 3
   
   Find when s(t) = 0:
   t² + 5t − 3 = 0
   t = (−5 ± √(25 + 12))/2 = (−5 ± √37)/2
   Taking t ≥ 0: t = (−5 + √37)/2 ≈ (−5 + 6.083)/2 ≈ 0.54 s
9. Show Solution
   (a) Direction changes when v = 0:
   v(t) = t² − 5t + 4 = (t − 1)(t − 4) = 0 ⇒ t = 1, t = 4
   Sign check: v(0) = 4 > 0; v(2) = −2 < 0; v(5) = 4 > 0
   Direction changes at t = 1 s (positive to negative) and t = 4 s (negative to positive).
   
   (b) Total distance from t = 0 to t = 5:
   s(t) = t³/3 − 5t²/2 + 4t (c = 0 since s(0) = 0)
   s(1) = 1/3 − 5/2 + 4 = 2/6 − 15/6 + 24/6 = 11/6
   s(4) = 64/3 − 40 + 16 = 64/3 − 24 = 64/3 − 72/3 = −8/3
   s(5) = 125/3 − 125/2 + 20 = 250/6 − 375/6 + 120/6 = −5/6
   
   Distance = |s(1)−s(0)| + |s(4)−s(1)| + |s(5)−s(4)|
   = |11/6 − 0| + |−8/3 − 11/6| + |−5/6 − (−8/3)|
   = 11/6 + |−16/6 − 11/6| + |−5/6 + 16/6|
   = 11/6 + 27/6 + 11/6 = 49/6 ≈ 8.17 m
10. Show Solution
    (a) Equations of motion:
    v(t) = ∫(−10) dt = −10t + c; v(0) = 15 ⇒ c = 15
    v(t) = −10t + 15
    s(t) = ∫(−10t + 15) dt = −5t² + 15t + k; s(0) = 2 ⇒ k = 2
    s(t) = −5t² + 15t + 2
    
    (b) Maximum height:
    v = 0: −10t + 15 = 0 ⇒ t = 1.5 s
    s(1.5) = −5(2.25) + 15(1.5) + 2 = −11.25 + 22.5 + 2 = 13.25 m
    
    (c) Total distance:
    Find when s = 0 (hits the ground):
    −5t² + 15t + 2 = 0 ⇒ 5t² − 15t − 2 = 0
    t = (15 + √(225 + 40))/10 = (15 + √265)/10 ≈ 3.127 s
    
    Split at t = 1.5 (when v = 0):
    Upward phase: s(0) = 2 → s(1.5) = 13.25, distance = 13.25 − 2 = 11.25 m
    Downward phase: s(1.5) = 13.25 → s(T) = 0, distance = 13.25 m
    Total distance = 11.25 + 13.25 = 24.5 m