SOLUTION: The position of a particle moving along the ๐‘ฅ-axis at any time ๐‘ก is given by ๐‘ (๐‘ก)=๐‘ก^3โˆ’11๐‘ก^2+24๐‘กโˆ’17. A) Find ๐‘ฃ(5) and ๐‘Ž(5). B) Is the particle spee

Algebra ->  Equations -> SOLUTION: The position of a particle moving along the ๐‘ฅ-axis at any time ๐‘ก is given by ๐‘ (๐‘ก)=๐‘ก^3โˆ’11๐‘ก^2+24๐‘กโˆ’17. A) Find ๐‘ฃ(5) and ๐‘Ž(5). B) Is the particle spee      Log On


   



Question 1147472: The position of a particle moving along the ๐‘ฅ-axis at any time ๐‘ก is given by
๐‘ (๐‘ก)=๐‘ก^3โˆ’11๐‘ก^2+24๐‘กโˆ’17.
A) Find ๐‘ฃ(5) and ๐‘Ž(5).
B) Is the particle speeding up
or slowing down at ๐‘ก = 5? Justify

Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.

(a) The velocity is the first derivative of the position function over time

        v(t) = %28ds%29%2F%28dt%29 = 3t^2 - 22t + 24.     (1)

     To find v(5),  substitute t= 5 into the formula (1)

     

    The acceleration is the second derivative of the position function over time

        a(t) = %28d%5E2s%29%2F%28dt%5E2%29 = 6t - 22.       (2)

     To find a(5),  substitute t= 5 into the formula (2).



(b)  At t= 5,  v(5) = -11,   a(5) = 8.


     So, the particle moves in negative direction and has positive acceleration.


     In the normal human language, it means that the particle DECELERATES (slowing down, moving in negative direction).