A particle moves along the x-axis so that its velocity is given by
v(t)= -(t+1)sin(t^2/2)
****at time t=0, the particle is at position x=1

a) Find the acceleation of the particle at time t=2. Is the speed of the particle increasing at t=2? Why or Why not?

b) Find all times t in the open interval 0,t,3 when the particle changes direction.Justify your answer.

c)Find the total distance traveled by the particle from time t=0 until time t=3

d) During the time interval 0<=t<=3, , what is the greatest distance between the particle and the origin?

I WOULD GREATLY APPRECIATE IT IF SOMEONE WOULD HELP ME OUT ON THIS PROBLEM AND GIVE ME THE CORRECT ANSWERS SO I CAN SEE IF MINE ARE CORRECT. I'M DESPERATE , PLEASE HELP.
- ALLEN

2. Originally Posted by sheven
A particle moves along the x-axis so that its velocity is given by
v(t)= -(t+1)sin(t^2/2)
****at time t=0, the particle is at position x=1

a) Find the acceleation of the particle at time t=2. Is the speed of the particle increasing at t=2? Why or Why not?
a)

Acceleration is the rate of change of speed, so:

$
a(t)=\frac{dv}{dt}=-\sin(t^2/2)-(t+1)t\cos(t^2/2)
$

where the derivative is found using a combination of the product rule and
the chain rule.

Substituting $t=2$ into this gives:

$a(2)\approx 1.588$

This is positive, and positive acceleration is the same thing as increasing
speed, so the speed is increasing.

RonL

3. Originally Posted by sheven
A particle moves along the x-axis so that its velocity is given by
v(t)= -(t+1)sin(t^2/2)
****at time t=0, the particle is at position x=1
b) Find all times t in the open interval 0,t,3 when the particle changes direction.Justify your answer.
b)

When the particle changes direction the velocity changes sign, and as
the velocity is a continuous function of time it must be zero at those
times that it changes direction.

If the velocity is zeros then:

$
v(t)=-(t+1)\sin(t^2/2)=0
$

For $t \in (0,3)$ (we ignore the end points as you can't change direction
at these points) this only occurs at the zeros of $\sin(t^2/2)$.
These are $t=\sqrt{2 n \pi},\ n=1,2, \dots$, and we now need to check
which values of $n$ give $t \in (0,3)$.

The only value of $n$ that gives a $t \in (0,3)$ is $1$ and the corresponding time
is $\approx 2.5066$

You will now have to check that this is in fact a point at which the velocity
changes sign rather than a point at which it is zero but does not change sign.
This means we need to check that the acceleration does not equal zero at
this time. Plugging $t=2.5066$ into the equation for the acceleration
found in part a) shows that the acceleration is indeed not zero, so the particle
changes direction at this time.

RonL

RonL

4. Originally Posted by sheven
A particle moves along the x-axis so that its velocity is given by
v(t)= -(t+1)sin(t^2/2)
****at time t=0, the particle is at position x=1
c)Find the total distance traveled by the particle from time t=0 until time t=3
c)

I find this slightly ambiguous does this mean the total distance travelled by
the partice or the distance of the particle at time $t=3$ from its position
at time $t=0$?

The position $s(t)$ at time $t$ is given by:

$
s(t)=\int_0^t v(\xi) d\xi +s(0) =\int_0^t -(\xi+1)\sin(\xi^2/2) d\xi +s(0)
$

The problem here is that this last integral is non-elementary (it can be written
as the sum of an elementary function and Fresnel's S function). This problem
can be overcome by integrating the functions numericaly.

Now the distance travelled by the particle between $t=0$ and $t=3$ is:

$
\rm{dist.}=\int_0^3 |v(\xi)| d\xi +s(0) =\int_0^3 |(\xi+1)\sin(\xi^2/2)| d\xi\approx 4.33
$

and the displacement of the final position from the starting position is:

$
\rm{disp.}=\int_0^3 v(\xi) d\xi =\int_0^3 -(\xi+1)\sin(\xi^2/2) d\xi\approx -2.20
$

d) During the time interval 0<=t<=3, , what is the greatest distance between the particle and the origin?
Again this can be done numericaly and is $\approx 2.27$.

RonL

6. Originally Posted by Vigo
I take the form

$
s(t)=\int_0^t -(\xi+1)\sin(\xi^2/2) d\xi +1
$

You want to find the point that it is furtherest form $0$ on $[ 0,3 ]$,
so differentiate and set the derivative to zero. Now:

$
\frac{d}{dt} s(t) =v(t)=-(t+1)\sin(t^2/2)
$
,

but we have already found the zeros of this in $[0,3]$, so we
need only check the end points and this zero at $t\approx 2.5066$
to find the extrema of $s(t)$.

Note that at this point $s \approx -2.27$, but the question asks
for the graetest distance so the sign is dropped.

RonL