Question

Please see the attachment!

Answer 1

How am I to do (a)?

Answer 2

I'm getting 0, 1.71, and 8.71, and is that right?

Answer 3

The "straight line" part of the question does not make sense. Graph the function. The particle is moving in a + direction approximately [0,2) and in a negative direction from (2,5) to go from positive to negative must go through a period of rest.

Answer 4

@hullsnipe Yes, I also noticed this which left me quite confused with how I was supposed to do this..

Answer 5

@hullsnipe straight line means it's moving in one dimension only.

Answer 6

a) first you need to find the first derivative of the function

Answer 7

@agent0smith Why using the 1st derivative? Can't I just find the zeros from the original equation?

Answer 8

s is the position function - when the object is at s=0 (ie the starting point) does that necessarily mean it is at rest?

Answer 9

The object is at rest when the velocity is zero; velocity is the derivative of position.

Answer 10

Ohh I see, thanks for clarifying that!