Acceleration as a Derivative

• v = ^ds/_dt
• a = ^dv/_dt = v ⋅ ^dv/_ds

Examples

1. A particle moves with acceleration a = -2v², where v is velocity.
   Initially, the particle is at 0 with v = 2
   1. Find an expression for v in terms of s
      • Express acceleration as a derivative with displacement
        • v ⋅ ^dv/_ds = -2v²
        • v ⋅ dv = -2v² ⋅ ds
      • Separate the variables and integrate
        • ∫^v/_v² ⋅ dv = ∫-2 ⋅ ds
        • ln v = -2s + c
      • Substitute the given information to find c:
        • ln 2 = 0 + c
        • ∴ c = ln 2
      • Rearrange the equation to make v the subject:
        • ln v = -2s + ln 2
        • ln v – ln 2 = -2s
        • ln(^v/₂) = -2s
        • ^v/₂ = e^-2s
        • ∴ v = 2e^-2s
   2. Find an expression for v in terms of t
      • Express acceleration as a derivative with time
        • ^dv/_dt = -2v²
        • v^-2 ⋅ dv = -2 ⋅ dt
        • ∫v^-2 ⋅ dv = ∫-2 ⋅ dt
      • Integrate the expression above:
        • -v^-1 = -2t + c
      • Substitute the given information to find c:
        • -(2)^-1 = -2(0) + c
        • ∴ c = -½
      • Rearrange the equation to make v the subject:
        • -v^-1 = -2t + -½
        • v^-1 = 2t + ½
        • v = ¹/_{2t + ½} = ²/_{4t + 1}

Variable Forces

• Bodies undergo variable acceleration due to the effect of variable forces, e.g. gravitational fields
• It is important to put a negative sign if the body is decelerating, e,g. when a body falls in a resistive medium
• Deriving an expression for force in terms of velocity and displacement:
  • F = ma
  • a = v^dv/_dx
  • F = mv^dv/_dx
• Deriving an expression for force in terms of work done:
  • For an object moving from x₁ to x₂, the change in kinetic energy or work done can be defined as:
    
    • ∴ ^dW/_dx = r
• Deriving an expression of power in terms of force and velocity:
  • P = ^dW/_dt = ^dW/_dx × ^dx/_dt = Fv

Examples

1. A car of mass 1200 kg is travelling on a straight horizontal road, with its engine working at a constant rate of 25 kW. Given that the resistance to motion of the car is proportional to the square of its velocity and that the greatest constant speed the car can maintain is 50 ms^-1, show that 125000 – v³ = 6000v² ^dv/_dt, where v ms^-1 is the velocity of the car when its displacement from a fixed point on the road is x meters
   • Find k:
     • F = kv²
     • ^P/_v = kv²
     • ²⁵⁰⁰⁰/₅₀ = k(50²)
     • ∴ k = 0.2
   • Write an equation that relates the forces:
     • Engine force = Work force + Resistive force
     • ^P/_v = mv^dv/_dx + kv²
     • ²⁵⁰⁰⁰/_v = 1200v^dv/_dx + ¹/₅v²
     • ¹²⁵⁰⁰⁰/_v = 6000v^dv/_dx + v²
     • 125000 = 6000v²^dv/_dx + v³
   • Find the distance covered by the car in increasing its speed from 30 ms^-1 to 45 ms^-1 by forming an integral to define displacement in terms of velocity
     • ^dv/_dx = ^{125000 – v³}/_6000v²
     • ^dx/_dv = ^6000v²/_{125000 – v³}
     • dx= (^6000v²/_{125000 – v³})dv
     • x = ∫(^6000v²/_{125000 – v³})dv
   • Substitute the known information: