• Mar 14th 2006, 05:46 PM
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?

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
• Mar 14th 2006, 11:23 PM
CaptainBlack
Quote:

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:

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

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

Substituting $\displaystyle t=2$ into this gives:

$\displaystyle a(2)\approx 1.588$

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

RonL
• Mar 14th 2006, 11:46 PM
CaptainBlack
Quote:

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:

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

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

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

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 $\displaystyle t=2.5066$ 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
• Mar 15th 2006, 04:51 AM
CaptainBlack
Quote:

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 $\displaystyle t=3$ from its position
at time $\displaystyle t=0$?

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

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

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 $\displaystyle t=0$ and $\displaystyle t=3$ is:

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

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

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

Quote:

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 $\displaystyle \approx 2.27$.

RonL
• Mar 15th 2006, 07:53 PM
Vigo
• Mar 15th 2006, 08:54 PM
CaptainBlack
Quote:

Originally Posted by Vigo

I take the form

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

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

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

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

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

RonL