Stanton D. answered • 02/26/21
Tutor to Pique Your Sciences Interest
Hi Iris E.,
I hope you realize that the standard way of interpreting your formula is (using parentheses for clarity):
s(t) = 2/(5t^5) - 2t^4 + 2t^3.
I'm pretty sure that what you meant instead is: s(t) = (2/5)*t^5 - 2t^4 + 2t^3, isn't it?
It sure makes a difference. And that's why you had PEMDAS drilled into you in grades 2, 3, 4, 5, 6, ...., isn't it? 'Nuff said. I'll suppose you meant the latter, as implied by the terms order.
So, velocity is the first derivative of position with respect to time.v = s'(t) = 5*(2/5)*t^4 - 4*2*t^3 + 3*2*t^2 and acceleration is the second derivative of position, or alternatively, the first derivative of velocity:
a = s"(t) = 4*5*(2/5)*t^3 - 3*4*2*t^2 + 2*3*2*t . You do take derivatives OK??
At rest? Solve for the roots of the velocity expression. You may immediately remove a factor of t^2, so that's a root of 0 of multiplicity 2. That leaves a quadratic form, hope you can handle that?
Now let's consider the motion of the particle. Both the roots of the quadratic are positive, so the portion between them will look like a usual quadratic, namely a parabola upwards. But between those 2 quadratic roots, a parabola dips down, doesn't it. So your velocity will be negative over that interval. The velocity will be positive between (0 not inclusive!) and the lesser quadratic root, and from the second quadratic root upwards. Can you predict what the expression would say for t<0? You SHOULD be able to reason, it's just a continuation of the parabolic shape, but distorted by that t^2 multiplier. I.e., the parabola upwards at large negative t is just "pinched down" to 0 velocity at t=0. Note that that is what is going on for v; the actual position s follows the original equation.
-- Cheers, --Mr. d.
