Class 11 · Derivations

Let a body have initial velocity u, constant acceleration a, and velocity v and displacement s after time t.

1. Start from the definition of acceleration, a = dv/dt, and integrate between 0 and t.
   ∫dv = a∫dt  ⇒  v − u = at  ⇒  v = u + at
2. Use v = ds/dt = u + at and integrate again.
   ∫ds = ∫(u + at)dt  ⇒  s = ut + ½at²
3. Write a = v dv/ds, so that a ds = v dv, and integrate from u to v.
   a·s = (v² − u²)/2  ⇒  v² = u² + 2as

A projectile is launched with speed u at an angle θ above the horizontal.

1. Resolve the velocity into components; gravity acts only vertically.
   u_x = u cos θ,   u_y = u sin θ,   a_x = 0,   a_y = −g
2. Time of flight: the body returns to the same level when y = 0.
   0 = (u sin θ)T − ½gT²  ⇒  T = 2u sin θ / g
3. Maximum height: at the top v_y = 0, so use v² = u² + 2as vertically.
   0 = u²sin²θ − 2gH  ⇒  H = u² sin²θ / 2g
4. Range: horizontal velocity is constant, so R = u_x × T.
   R = u cos θ · (2u sin θ / g)  ⇒  R = u² sin 2θ / g
5. R is maximum when sin 2θ = 1, i.e. θ = 45°.
   R_max = u²/g

A net force F acts on a body of mass m along the x-axis, changing its speed from v₁ to v₂.

1. By Newton's second law, with a = dv/dt = v dv/dx:
   F = m a = m v (dv/dx)
2. Work done is the integral of F over displacement.
   W = ∫F dx = ∫_v1^v₂ m v dv
3. Integrate:
   W = ½mv₂² − ½mv₁² = ΔK
4. So the net work done on a body equals the change in its kinetic energy.

Let a mass m be projected from the surface of a planet of mass M and radius R.

1. Gravitational potential energy at the surface:
   U = −GMm / R
2. To escape to infinity, the total energy must at least be zero (the body just reaches infinity at rest).
   ½mv_e² − GMm/R = 0
3. Solve for v_e:
   v_e = √(2GM / R)
4. Using GM = gR²:
   v_e = √(2gR) = √(2 × 9.8 × 6.4×10⁶) ≈ 11.2 km/s

Consider steady, incompressible, non-viscous flow in a pipe from section 1 (P₁, A₁, v₁, h₁) to section 2 (P₂, A₂, v₂, h₂).

1. By continuity, the same volume ΔV flows through both sections in time Δt.
   ΔV = A₁v₁Δt = A₂v₂Δt
2. Net work done by the pressure forces on this fluid:
   W = (P₁ − P₂) ΔV
3. Change in kinetic and potential energy of the mass ρΔV:
   ΔK = ½ρΔV (v₂² − v₁²),   ΔU = ρΔV g (h₂ − h₁)
4. Apply the work–energy theorem, W = ΔK + ΔU, and rearrange:
   P + ½ρv² + ρgh = constant

Take one mole of an ideal gas and heat it by dT.

1. At constant volume no work is done, so all the heat raises internal energy.
   dQ = C_v dT = dU
2. At constant pressure, heat raises internal energy and also does work.
   C_p dT = dU + P dV
3. For an ideal gas, U depends only on T, so dU = C_v dT in both cases.
   C_p dT = C_v dT + P dV
4. From PV = RT at constant P, we get P dV = R dT. Substitute and cancel dT.
   C_p − C_v = R

A mass m attached to a spring of constant k is displaced by x from equilibrium.

1. Restoring force (Hooke's law) and Newton's second law:
   F = −kx  ⇒  m (d²x/dt²) = −kx
2. Rewrite in the standard SHM form with ω² = k/m:
   d²x/dt² = −ω²x
3. The solution is x = A cos(ωt + φ), which repeats when ωT = 2π.
   T = 2π/ω  ⇒  T = 2π √(m/k)