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A particle's acceleration along a straight line is given by #a(t)=48t^2+2t+6# . It's initial velocity is equal to -3cm/s and its initial position is 1 cm. Find its position function s(t). Answer is #s(t)=4t^4+1/3t^3+3t^2-3t+1# but I can't it figure out?

#s(t)=4t^4+1/3t^3+3t^2-3t+1#

Answer 1

#"See explanation"#

#a = {dv}/dt# #=> v = int a(t) dt# #= 16 t^3 + t^2 + 6 t + C# #v(0) = v_0 = -3 => C = -3# #=> v = 16 t^3 + t^2 + 6 t - 3#

#v = {ds}/dt" (v = velocity)# #=> s = int v(t) dt# #= 4 t^4 + t^3/3 + 3 t^2 - 3 t + C# #s(0) = s_0 = 1 => C = 1# #=> s(t) = 4 t^4 + t^3/3 + 3 t^2 - 3 t + 1#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.