Question

The function ​v(t)equals=t cubed minus 10 t squared plus 21 tt3−10t2+21t​, ​[0,99​], is the velocity in​ m/sec of a particle moving along the​ x-axis. Complete parts​ (a) through​ (c). a. Determine when the motion is in the positive direction and when it is in the negative direction. b. Find the displacement over the given interval. c. Find the distance traveled over the given interval.

Answer 1

NOTE: I suppose there was an error when typing the Interval.

Graphing the graph up to 99 for example for point A, implies a 10-fold perioricity and for point B almost 10 integrals! That is a little expensive!

Which implies an extremely costly graph to count the positive and negative points.

I will start by performing the procedure for a shorter interval, in case it is necessary and implicit

Extensive interval, I think the procedure is contingent and it is simply a matter of extrapolating it.

 

So things,

The function:

$v (t) = t ^ 3 - 10t ^ 2 + 21t, [0.9]$

a) $t \in (0.3) \Rightarrow$ Positive address & $t \in (7.9)$

$t \in (3.7) \Rightarrow$ Negative address

b) $Displacement = \int \limit ^ 9_0 (t ^ 3-10 ^ t2 + 21) dt = \frac {t ^ 4} {4} - \frac {10 ^ 3} {3} + \frac {21t^2} {2} \Big| _0 ^ 9$

Displacement = 60.75m

c) $Distance = \int \limit ^ 0_3 (t ^ 3-19t ^ 2 + 21t) dt + \int \limit ^ 7_3 (-t ^ 3 + 10t ^ 2-21t) dt + \int \limit ^ 9_7 (t ^ 3-19t ^ 2 + 21t) dt$

Distance = 167.41m