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# Physics Problem! Kinematics and position?

The motion of a particle is defined by the relation

x=2t^3−15t^2+24t+4 , where x and t are expressed in m and s. Determine where the velocity is zero and the position and total distance traveled when acceleration is zero.

This problem is really annoying! I have the answer to all the answers but the last one but the total distance traveled! Pls help!  The velocity equation is the 1st derivative of the distance equation.

v = 6t^2 – 30t + 24

t = [30 Â± â(900 – 576) Ã· 6

t = [30 Â± 18] Ã· 6

t = 8 seconds and 2 seconds

Now we can determine 2 positions.

x= 2 * 8^3 â 15 * 8^2 + 24 * 8 + 4 = 1024 – 960 + 192 + 4 = 260 m

x= 2 * 2^3 â 15 * 2^2 + 24 * 2 + 4 = 16 – 60 + 48 + 4 = 8 m

The acceleration equation is the 1st derivative of the velocity equation.

v = 6t^2 – 30t + 24

a = 12 t – 30 = 0

t = 2.5 seconds

x= 2 * 2.5^3 â 15 * 2.5^2 + 24 * 2.5 + 4 = 31.25 – 93.75 + 60 + 4 = 1.5 m v = x' = 6t^2 - 30t + 24

a = v' = 12*t - 30

v = 0 = 6t^2 - 30t + 24 = 6*(t^2 - 5t + 4) = 6*(t - 4)(t - 1)

0 = 6*(t-4)*(t-1) ==> t = 4 and t = 1

Plug those numbers back into the x(t) equation to find the position.

a = 0 = 12*t - 30 ==> t = 2.5

Total distance traveled = h(2.5) - h(0)