Question

PLEASE HELP a particle moves so that x, its distance from the origin at time t, t is greater then or equal to 0 is given by x(t)=cos^3(t). the first interval in which the particle is moving to the right it
1.)0< t 10 years ago

Answer 1

Differentiate x(t) wrt t to get v(t)

Answer 2

you mean to find the velocity?

Answer 3

yes. please write down the expression for the velocity for me.

Answer 4

3(cos(t))^2(-sin(t))

Answer 5

If v(t) is +ve then the object is going forward(let's consider it right..since its 1D motion) and backwards if its negative.
Evaluate your v(t) expression for all the given options.

Answer 6

i'm sorry because i really font know how to do this, would that mena i set the v(t) to equal zero to find the roots?

Answer 7

I'll do the first one for you.
When t is from 0 to pi/2
So lets take t = pi/4.
\[\Large v(\pi/4) = - 3 \cos^2(\frac{\pi}{4})\sin (\frac{\pi}{4}) <0\]
So its not A.

Answer 8

OH! Aright one moment ill do the next one

Answer 9

sure :)

Answer 10

On a side note, you can do this manually too, that was just one way I told you.
Notice v(t) more carefully.
\[\Large v(t) = - 3 \cos^2(t) \sin(t)\]
Observe that cos^2(t) is always +ve..so if we somehow get sin(t)<0 then v(t) would be +ve.
So it boils down to ..in which quadrant is sine negative ?

Answer 11

\[3\cos(3\pi/4)^2(-\sin(3\pi/4))<0\]

Answer 12

After you get that answer, then you can go for the first method to cross check your answer..just pickup a number from the range and substitute to check whether its +ve or not.

Answer 13

sin is negative in three and four i think

Answer 14

i'm not sure what you mean

Answer 15

Yep..so " the first interval " phrase in the question will help your decide between the 3rd and the 4th quadrant

Answer 16

Since 3rd quadrant is the first interval where sine is negative, therefore t is from pi to 3pi/2.
Now if you want to check your answer then choose a number between [pi,3pi/2] ..say 3pi/4.
\[\Large v(\frac{3 \pi}{4}) = - 3 \cos^2(\frac{3\pi}{4})\sin(\frac{3\pi}{4}) = -3*(\frac{-1}{\sqrt2})*(\frac{1}{\sqrt2})>0\]

Answer 17

so our answer is correct and verified.

Answer 18

I'm not sure this is correct or even relevant but when i plug in v(t) into my graphing calculator then the first two waves are below the x axis in the third quadrant is that significant, or an i graphing the wrong equation?

Answer 19

you don't need to check it, it was an additional step.
the question is just asking for the first interval where sine is negative (so that v(t) is positive) and its the 3rd quadrant.

Answer 20

ok and the first interval would be 3pi/2 < t <2pi

Answer 21

3rd quadrant means theta ranges between:
\[\Huge [\pi,\frac{3\pi}{2}]\]

Answer 22

right, becasue when i plug in pi/2 into the v(t) equation the answer is 0

Answer 23

THANK YOU FOR YOUR TIME @DLS