The position

The position of a body at any time T is given by the displacement function S=t3-2t2-4t-8. Find its acceleration at each instant time when the velocity is zero.

Result

a =  8

Solution:

s=t32t24t8 v=ds/dt=3t24t4 a=dv/dt=6t4 v=0   3 t24 t4=0 3t24t4=0  a=3;b=4;c=4 D=b24ac=4243(4)=64 D>0  t1,2=b±D2a=4±646 t1,2=4±86 t1,2=0.66666667±1.3333333333333 t1=2 t2=0.66666666666667   Factored form of the equation:  3(t2)(t+0.66666666666667)=0  t>0 a=6 t14=6 24=8s=t^3-2t^2-4t-8 \ \\ v=ds/dt=3t^2 - 4t - 4 \ \\ a=dv/dt=6t - 4 \ \\ v=0 \ \\ \ \\ \ \\ 3 \cdot \ t^2 - 4 \cdot \ t - 4=0 \ \\ 3t^2 -4t -4=0 \ \\ \ \\ a=3; b=-4; c=-4 \ \\ D=b^2 - 4ac=4^2 - 4\cdot 3 \cdot (-4)=64 \ \\ D>0 \ \\ \ \\ t_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 4 \pm \sqrt{ 64 } }{ 6 } \ \\ t_{1,2}=\dfrac{ 4 \pm 8 }{ 6 } \ \\ t_{1,2}=0.66666667 \pm 1.3333333333333 \ \\ t_{1}=2 \ \\ t_{2}=-0.66666666666667 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 3 (t -2) (t +0.66666666666667)=0 \ \\ \ \\ t>0 \ \\ a=6 \cdot \ t_{1}-4=6 \cdot \ 2-4=8

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