i cant figure out how to start this........?

Question

An object moves along the x-axis in a horizontal path in such a way that its poistion at time "t" is given by
An object moves along the x-axis in a horizontal path in such a way that its poistion at time "t" is given by s(t)=(1-t^3)/(1+t^4) for t >(or=to)0

a.) Find the velocity, v(t), &determine the time interval(s), accurate to 6 significant figures, during which the particle is moving to the right.

b.)Find the acceleration,a(t), and use it to find the maximum speed, accurate to 6 significant figures, of the particle & the time, accurate to 6 significant figures, at which the maximum speed is attained.

Answer 1

from your explanation, it looks like s(t) is a position function....

so, to find the velocity, just find its first derivative...

s(t)=(1-t^3)/(1+t^4)

s'(t) = v(t) = [(1+t^4)(-3t^2) - (1-t^3)(4t^3)] / (1+t^4)^2

to find t where the object moves to the right. just equate v(t) = 0 or find where v(t) doesn't exist...

in this case, its either (1+t^4)^2 = 0 or (1+t^4)(-3t^2) - (1-t^3)(4t^3) = 0

so...

-3t^2-3t^6 - 4t^3 + 4t^6 = 0

t^6 - 4t^3 - 3t^2 = 0
t^2(t^4-4t-3) = 0

t can be 0 or the solution, t^4-4t-3 = 0(use a graphing application/calculator or Newton's Method to find the zeroes...)
then, try it for the s(t)...

to find a(t), get the derivative of v(t), and then set a(t) = 0... solve for the t's where it a(t) = 0

then set it on v(t)

Answer 2

Differentiate the s(t) expression with respect to t to get v(t) velocity in terns of t.

Put v(t) = 0, or rather the numerator of v(t) =0 to find the timeat which it turns back. this might involve solving a quadratic/cubic (whatever, i didn't try the question, so I can't really say.) You know the quotient rule for differentiation, right?

b> Further differentiate v(t) with respect to time to get accelaration expression. Equate it to 0 to get points of maxima for velocity, and the put those values in v(t) to get maximum velocity at those times.