CAN ANYONE HELP WITH CALCULUS RIGHT THIS VERY MOMENT?

Question

sillybilly123 · Answer

We have displacement: $s(t) = t^3  -  12t^2+45t+4$

Differentiate to get velocity $\dot s = v = 3t^2  -  24t+45$

So at $t = 0$, we have $v = 45$ ft/s, a positive value.  We want to know when the particle changes direction, ie when the velocity becomes negative.

At some point between moving in a positive direction and moving in a negative direction, the particle must stop moving at all.  We are looking for those points, if any, where $v(t) = 0$.

So we factor:

$v =3 (t^2  -  8t+15)$

$=3 (t- 3)(t - 5)$

So $v(t) = 0 \implies t = 3,5$, ie the particle is stationary at $t = 3,5$.

Upon inspection of $v(t)  =3 (t- 3)(t - 5)$, you can see that $v(t)$ is negative for $t \in (3,5)$ because the $(t - 5)$ term is negative whereas the $(t - 3)$ term is positive.

So the particle first reverses direction at $t = 3$.

At that point in time......

Displacement: $s(3) = 58$ ft

Acceleration.....differentiate again:

$\ddot s = \dot v = a = 6t - 24$

$ a(3) = -6$ ft/sec/sec

iwanttogotostanford · Answer

@sillybilly123 thank you sir! is that it? :-)

iwanttogotostanford · Answer

i can now use it to study and guide me through the rest!

iwanttogotostanford · Answer

@sillybilly123 ?

sillybilly123 · Answer

yeah, that's it :)

iwanttogotostanford · Answer

thank you so much! could you possibly help me with another one please?

iwanttogotostanford · Answer

@sillybilly123