# Homework Help: Position velocity equation free response

1. May 5, 2013

### syeh

1. The problem statement, all variables and given/known data
A partcile starts att he point (1,0) and t=0 and moves along the x-axis so that at time t≥0 its velocity v(t) is given by v(t)= $\frac{1+ t}{(1+t^2)}$

A) Determine the maximum velocity of the particle.

B) Find an expression for the position s(t) of the particle at time t.

C) What is the limiting value of the veolicty as t increases without bound?

D) Determine for which values of t, if any, the particle reaches the point (101,0)

2. Relevant equations

3. The attempt at a solution

A) To find the maximum velocity:
Checked endpoints: t=0
v(0)=1

Where v'(t)=0: (using chain rule:
v'(t)= $\frac{(1+t^2)(1)-(t)(2t)}{(1+t^2)^2}$
=$\frac{1+ t^2 - 2t^2}{(1+t^2)^2}$
0=1+ t^2 - 2t^2
t=0.577
V(.577)=1.433

Where v'(t)=∅
none

So the max. velocity is 1.433

The answers said it was 1.5, but that might just be from rounding

B) I dont know how to find an expression of the position. it starts at (1,0) which is an odd place to start. how to i put that into the equation?

The answer says s(t) = 1+ t + ln√(1+t^2)

How did they get that?!?!?!?

C) So I guess limiting value is asking if its approaching a number? Like a limit? How do i find that?

2. May 5, 2013

### Simon Bridge

$$v(t)=\frac{1+t}{1-t^2}$$... by the quotient rule, turning points where: $$\frac{dv}{dt}=0$$ ... I don't know why you cited the chain rule here, use the quotient rule.
Anyway, I don't think this step was correct. Review: $$\left ( \frac{f}{g} \right )^\prime = \frac{gf^\prime - fg^\prime}{g^2}$$ .. what are f and g and f' and g' in your case?

3. May 5, 2013

### Alcubierre

(A) seems right to me. As for (B), what happens when you integrate velocity? For (C) it's asking for the limit of the velocity if t is boundless, so what does that mean?

EDIT: Actually, now that I look at your solution for (A), you are forgetting a 't' in the numerator when you apply the quotient rule. Look closely at it.