# Thread: Position, Velocity, Acceleration problem

1. ## Position, Velocity, Acceleration problem

A particle moves along the $x-axis$ so that at any time $t \geq 1$ its accelaration is given by $a(t)= \frac{1}{t}$. At $t=1$, the velocity of the particle is $v(1)=-2$ and its position is $x(1)=4$.
a) Find the velocity $v(t)$ for $t \geq 1$.
b) Find the position $x(t)$ for $t \geq 1$.
c) What is the position of the particle when it is farthest left?

I answered parts (a) and (b) (unsure if they're correct):
a) $v(t)=\ln {t}-2$
b) $x(t)=t\ln {t}-3t$

However I cannot answer part (c), here's what i have done(I have a bad feeling I am totally wrong):
$\int_1^\infty (\ln {t}-2) \, dt = \infty$

Can anyone help?

2. Originally Posted by xxsteelxx
A particle moves along the $x-axis$ so that at any time $t \geq 1$ its accelaration is given by $a(t)= \frac{1}{t}$. At $t=1$, the velocity of the particle is $v(1)=-2$ and its position is $x(1)=4$.
a) Find the velocity $v(t)$ for $t \geq 1$.
b) Find the position $x(t)$ for $t \geq 1$.
c) What is the position of the particle when it is farthest left?

I answered parts (a) and (b) (unsure if they're correct):
a) $v(t)=\ln {t}-2$
b) $x(t)=t\ln {t}-3t$

However I cannot answer part (c), here's what i have done(I have a bad feeling I am totally wrong):
$\int_1^\infty (\ln {t}-2) \, dt = \infty$

Can anyone help?
a) is correct
b) I think you are missing a constant

for the last part here is a hint when does the velocity (part a) change from negative( going left) to positive (going right).

3. Originally Posted by xxsteelxx
A particle moves along the $x-axis$ so that at any time $t \geq 1$ its accelaration is given by $a(t)= \frac{1}{t}$. At $t=1$, the velocity of the particle is $v(1)=-2$ and its position is $x(1)=4$.
a) Find the velocity $v(t)$ for $t \geq 1$.
b) Find the position $x(t)$ for $t \geq 1$.
c) What is the position of the particle when it is farthest left?

I answered parts (a) and (b) (unsure if they're correct):
a) $v(t)=\ln {t}-2$
b) $x(t)=t\ln {t}-3t$
Is that "3" a typo? the integral of ln(t)- 2 is t ln(t)- 2t+ C. For x(1)= 4, you must have C= 6.

However I cannot answer part (c), here's what i have done(I have a bad feeling I am totally wrong):
$\int_1^\infty (\ln {t}-2) \, dt = \infty$

Can anyone help?