Particle moves in straight line and has acceleration given by a(t) = 6t^2 + 1

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Suresh Chand · Accepted Answer

Solution Given a(t)  = 6t2 + 1 Initial Velocity (u) = 4m/sec initial displacement (s(0)) = 5cm Position function (s(t)) = ? We know v'(t) = a(t) = 6t2 + 1 Integrating we get, (v(t) = 6frac{t^3}{3} + t + C) (v(t) = 2t^3 + t + C) As we know, v(0) = 4m/sec then c = 4 (v(t) = 2t^3 + t + 4) Again s(t) = ∫v(t) dt s(t) = ∫(2t3 + t + 4) dt (s(t) = frac{t^4}{2} + frac{t^2}{2} + 4t + D) Since s(0) = 5 so D = 5 Therefore position function is given by (s(t) = frac{t^4}{2} + frac{t^2}{2} + 4t + 5)

A particle moves in a straight line and has acceleration given by a(t) = 6t² + 1. Its initial velocity is 4m/sec and its initial displacement is s(0) = 5cm. Find its position function s(t).

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A particle moves in a straight line and has acceleration given by a(t) = 6t2 + 1. Its initial velocity is 4m/sec and its initial displacement is s(0) = 5cm. Find its position function s(t).

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A particle moves in a straight line and has acceleration given by a(t) = 6t² + 1. Its initial velocity is 4m/sec and its initial displacement is s(0) = 5cm. Find its position function s(t).