Cal Question?

Question

A particle is moving so that at time t, the distance is given by the function:

s(t) = -1/3(t)^3 - 5/2(t)^2 + 6(t)

At what time t greater than zero is the speed equal to zero

Answer 1

The speed is given by

s'(t) = -t^2 - 5t + 6.

Factor this. First, factor out a -:

s'(t) = -(t^2 + 5t - 6).

Then factor the rest:

s'(t) = -(t + 6)(t - 1).

Then s'(t) = 0 when t=-6 or t=1. Therefore your answer is t=1.

Answer 2

let s = -1/3(t)^3 - 5/2(t)^2 + 6(t)

If you differentiate the above equation, you will get the speed, hence ;

s'(t) = -t^2 - 5t + 6

When speed or s'(t) = 0,
hence

0 = -t^2 - 5t + 6
solve for t with (-b (+ or -) root (b^2 - 4ac)) / 2a

you will get t is either 1 or -6

t is time and wouldn't be -ve, hence the answer shall be 1 second

Answer 3

Man, I thought it said cat question. My eyes are getting bad. Who's going to want to do your math homework for you?