# Mathematics-Kinematics help

1. Jan 24, 2009

### icystrike

The question goes like this:
A body moves in a straight line from a fixed point O . Its distance from O, s meters , is given by s=t-$$\frac{1}{9}$$t³ , where t is the time in seconds after passing through O. Fubd

(a)the time when the body returns to O.

I have done this question by plugging s=0 and i get t=3

(b)the velocity at this instant.

I found the $$\frac{ds}{dt}$$ and i plug in t=3 and i got -2m/s

(c)the value of t when the body is instantaneously at rest.

I assumed $$\frac{ds}{dt}$$=0 and i've gotten myself t=$$\sqrt{3}$$

(d) the distance moved by the body in the 2nd second.

I dont know what is the question asking , but the answer should be 0.304.

2. Jan 24, 2009

### CompuChip

Re: Mathematics-Kinematics

Your approach for the first three looks correct, although for b and c I didn't check the numbers (so assuming that you did the math correctly as well you should have found the correct answers).

For d, I suppose you would just argue like: the first second is the time between t = 0 and 1, the 2nd second is between t = 1 and t = 2, etc. - then find the distance that is travelled between those times.

3. Jan 24, 2009

### icystrike

Re: Mathematics-Kinematics

Thanks JIANKAI!!
Thank you chip (:

I know how its works le.

Let f(x)=t-$$\frac{1}{9}$$t³

f(root3)-f(1) + f(root3)-f(2)

4. Jan 24, 2009

### CompuChip

Re: Mathematics-Kinematics

What or who is Jiankai?

And are you sure about that answer? The signs look a bit off to me. How did you get that?

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