# 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).?

May 31, 2016

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 $\pm 6 c m \text{/} {\sec}^{2}$

#### Explanation:

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

$\left(- 1\right) \cdot \left(- 3\right) = 3$ and $\left(- 1\right) + \left(- 3\right) = - 4$

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