Question

a particle moves along the y-axis so that its velocity at any time t>(or equal to)0 is given by v(t)= tcost. At time t=0, the position of the particle is y=3.
a) for what values of t, 0<(or equal to)t<(or equal to)5, is the particle moving upward?

Answer 1

t is always positive and
\[\cos(t)>0\] on
\[(0,\frac{\pi}{2})\] and negative on
\[(\frac{\pi}{2},\frac{3\pi}{2})\]

Answer 2

So, basically you need to find displacement. Remember:
\[r(t)=\int v(t) dt\]
So, from that:
\[\int t\cos (t)dt=\sin (t) + \cos (t) + C\]
They give us an initial condition:
r(0) = 3
So, we need to find when this is possible:
3= sin(0) + cos (0) + C
C=2
Now we have our displacement:
\[r(t) = \sin (t) + \cos (t) + 2\]
Now we are ready for the question.
For what values of
\[0\le t \le 5\]
For this we need to know if r(t) will be positive, probably the easiest route would be to graph and see what it looks like, or you can plug in worthwhile numbers and see what turns up.

Answer 3

oh up until 5, so have some more positive part on
\[(\frac{3\pi}{2},\frac{5\pi}{2})\]

Answer 4

@agreene did i totally mess this up? i thought if the velocity is positive you are going up.

Answer 5

In a sense, yes.
But that isn't proof that you aren't in an odd space.

Answer 6

I'm confused, I don't think we went that far into our lesson. The question comes from the notes for acceleration....so where did displacement come in?

Answer 7

maybe there is another part to this question, but first part just says velocity is
\[t\cos(t)\] so if it is positive particle is going up and if it is negative it is going down

Answer 8

doesn't ask "where are you?" just "what direction are you going?"

Answer 9

For instance, this displacement never crosses the x axis, so it never changes direction. it is positively moving the whole time, but with variable speed.

Answer 10

hold on. the velocity is negative some of the time. that means it does go down. maybe not below the x axis, but it does go downward

Answer 11

A negative velocity is only indicative of a change in velocity, for those sections negative
V_f - V_i < 0
This doesn't imply a change in direction

Answer 12

i think i am going to disagree. negative velocity means you are going backwards.

Answer 13

In some reference frames yes, but in this case... the reference frame in undefined, so we cannot assume we are in that space.

Answer 14

^ Most reference frames.

Answer 15

just like saying throw a rock in the air, but instead of having the equation be
\[h(t)=h_0+v_0t-16t^2\] we have
\[h'(t)=-32t+v_0\] we have
\[v(t)=t\cos(t)\] if it is positive thing is going up on the y - axis and if it negative it is going down

Answer 16

you are told what the velocity is. if the velocity is positive, particle is going up if it is negative, it is going down.

Answer 17

i am not sure what "reference frame is undefined" means

Answer 18

I completly understand what you are saying, and I agree that in most cases you are right, however. This time, it doesn't really work that way.

t cos (t) is only describing the velocity in the y direction, changes in velocity indicate -x +x movement within that reference frame, and not a change in y values.

Answer 19

okay so the 3nd part asks
b) write an expression for the acceleration of the particle in terms of t

Answer 20

take the derivative using the product rule

Answer 21

v'(t) = a(t)
d/dt (tcost) = cos(t)-sin(t)

Answer 22

\[v'(t)=\cos(t)-t\sin(t)\]

Answer 23

yes, satellite is correct, I forgot my silly t

Answer 24

okay that was understandable. but im still confused on part A)

Answer 25

i am sticking to my answer (which may be wrong) that if the velocity is positive you are headed up, and in the velocity is negative you are headed down. so question is really
"for which t on [0,5] is
\[t\cos(t)>0\]?"