he motion of a particle is described by the equation x= 20cm+(4cm/ s^2)t^2 To find:A) Find the displacement of the particle in the INTERVAL t=2 s and t’=5 s. B) Find the average velocity in this time interval C) Find instantaneous velocity at time t=2Solution:From given, we have,An equation that represents the motion of a particle,x = 20cm+(4cm/ s^2)t^2  ⇒ x = a + bt²The general form of equation of motion is given by,x = x₀ + V₀t + At²/2where, A = ACCELERATION and V₀ = initial velocityComparing the given equation with the standard equation, we have,x₀ = 20, V₀ = 0 and A = 2B = 2(4) = 8A) The displacement of the particle in the interval t=2 s and t’=5 s. ΔS = x_{T2} - x_{t1} = a + b_{t2}² - a - b_{t1}² = b[{t2}² - {t1}²] ΔS = 4 [5² - 2²] = 4 [25 - 4] = 4 [21] = 84 ∴ The displacement of the particle in the interval t=2 s and t’=5 s is 84 cm.B) The average velocity in this time interval V_{a} = ΔS/Δt = ΔS / [t2 - t1] = 84/[5 - 2] = 84/3∴ V_{a} = 28∴ The average velocity in this time interval is 28 cm/sC) Find instantaneous velocity at time t=2The  equation of motion of a particleV = V₀ + Atwhere, V₀ = 0 and A =  2b = 2 × 4 = 8 cm/s²The instantaneous velocity at t = 2Vi = At = 8 × 2 = 16Therefore, the instantaneous velocity at time t=2 is 16 cm/s