The position of an object is given by $${x(t)= 2.17+4.80t^2-0.100t^6}$$
Would I have to take the derivative of that in order to find velocity?
But from there how would I find the position and acceleration the instant velocity is zero? For time I thought about setting v to zero and solving to find the time but I don't know how that would really help me.
Would acceleration just be the derivative of v? Is there an exact way to find acceleration?
The position of an object is given by
$${x(t)= 2.17+4.80t^2-0.100t^6}$$
Thus velocity of the object is given by x'(t) which is:
$$x'(t)=v(t) = 9.60t-0.600t^5$$
Since velocity is 0, we can solve for time:
$$0= 9.60t-0.600t^5$$
$$ t= -2,0,2$$
Since time can't be negative, and we probably want the positive t -value, let's use t = 2.
We then sub t = 2 into x(t) and a(t).
We can get a(t) by using the identity v'(t)=a(t)
$$v'(t)=a(t)=9.6-3t^4$$
Solving for t =2:
$$a(2)=9.6-3*2^4=-38.4$$
$${x(2)= 2.17+4.80*2^2-0.100*2^6=14.97}$$
Therefore, position and acceleration when velocity is equal to 0, is 14.97 $units$ and -39.4 $units/s^2$