Question

Calculus:
A particle moves along the x-axis such that its position, x, at any time, t, where t>0, is given by the equation x(t)=sin(t^3).
a) Find the speed of the particle when t = pi/4 seconds. (I got this part - 1.309 is my answer)
b) Find the acceleration of the particle when t = pi/4 seconds. (I got this too- .911)

Now I got confused.
c) Is the particle speeding up or slowing down at t = pi/4 seconds?
d) Which direction is the particle moving at t = pi/4 seconds?
e) When is the first time, t>0, that the particle changes direction?

Any help is appreciated!

Answer 1

at any given time t, particle speeds up if v(t) and a(t) have a same sign
and particle slows down if v(t) and a(t) have positive sign

Answer 2

Okay that makes sense. And for d), would it be to the right because v(t) is positive?

Answer 3

assuming x+ is to the right. then if v(t) > 0 it's moving to the right
if v(t) < 0, it's moving to the left

Answer 4

so yeah

Answer 5

Alright, so then how would I go about e? That's the most confusing part to me.

Answer 6

what happens to the sign of v(t) when the particle changes direction?

Answer 7

The sign would change, too.

Answer 8

correct. So you just need to find when that first happens

Answer 9

So... do I set v(t) = 0? And then test points around those values to see where it changes?

Answer 10

yes

Answer 11

particularly where it first changes sign

Answer 12

Okay..thank you so much!