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If a particle moves according to the equation #s = t3 – 6t2 + 9t – 5#, how do you find the acceleration when the velocity is #0# (#s# is in cm).?

Answer 1

Take the derivative to find the expression for the velocity of the particle. Take the derivative again to find the expression for the acceleration. Set the velocity expression equal to zero and solve the resulting quadratic for s. Plug this value into the expression for the acceleration. You should get #pm 6cm"/"sec^2#

#s = t^3 – 6t^2 + 9t – 5#

#v={ds}/{dt} = 3t^2 – 12t + 9#

#a={dv}/{dt}={d^2s}/{dt^2} = 6t – 12#

#v= 3t^2 – 12t + 9=0#

#3(t^2 – 4t + 3)=0#

#(-1)*(-3)=3# and #(-1)+(-3)=-4#

so #x-3#, and #x-1# are factors

#3(t – 3)(t - 1)=0#

we have roots #t_1=1# and #t_2=3#

Plug these into #a#

#a_1 = 6(1) – 12=-6#

#a_2 = 6(3) – 12=6#

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Answer 2

Piper Hoffman

The acceleration equation is found by twice differentiating the displacement equation, which yields the acceleration when the velocity is zero.

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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.