The velocity of a particle is v = t³ − 6t² + 8t + 1. If t₁ and t₂ are the two instants at which acceleration is zero, then v(t₁)+v(t₂) is:

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G1 Physics · Accepted Answer

a = dv/dt = 3t² − 12t + 8. Let the roots of a=0 be t₁ and t₂. Then t₁+t₂ = 4, t₁t₂ = 8/3. Now v(t₁)+v(t₂) = (t₁³+t₂³) − 6(t₁²+t₂²) + 8(t₁+t₂) + 2. Using t₁²+t₂² = 16−16/3 = 32/3, t₁³+t₂³ = 4³ − 3(8/3)(4) = 32, we get 32 − 64 + 32 + 2 = 2 m s⁻¹.