# Thread: How do I do this particle question? (derivatives)

1. ## How do I do this particle question? (derivatives)

Hi, can someone help me with this?

A particle moves according to a law of motion s = f(t), t>0, where t is seconds and s is feet

for $f(t) = te^{-t/2}$,

a. Find the velocity at time t

so I know to find the derivative, which I got:

$te^{-t/2}(-\frac{1}{2}lnte + \frac{e}{te})$
^pretty sure that's wrong.. can anyone help me with that?

b. When is the particle at rest?

c. When is the particle moving in the positive direction?

d. What is the total distance traveled in the first 8s?

e. Find the acceleration at time t after 3s

f. When is it speeding up? When is it slowing down?

How do I do these as well? With the function I'm using, I am pretty clueless.

Thanks!

2. Originally Posted by coldfire
Hi, can someone help me with this?

A particle moves according to a law of motion s = f(t), t>0, where t is seconds and s is feet

for $f(t) = te^{-t/2}$,

a. Find the velocity at time t

so I know to find the derivative, which I got:

$te^{-t/2}(-\frac{1}{2}lnte + \frac{e}{te})$
^pretty sure that's wrong.. can anyone help me with that?

b. When is the particle at rest?

c. When is the particle moving in the positive direction?

d. What is the total distance traveled in the first 8s?

e. Find the acceleration at time t after 3s

f. When is it speeding up? When is it slowing down?

How do I do these as well? With the function I'm using, I am pretty clueless.

Thanks!
$f(x)=xe^{\frac{-x}{2}}\Rightarrow{f'(x)}=e^{\frac{-x}{2}}-\frac{x}{2}e^{\frac{-x}{2}}=\left(1-\frac{x}{2}\right)e^{\frac{-x}{2}}$

So I will tell you what you need to do..and you do it

b. particle is at rest when $v(t)=s'(t)=0$ Now remember that $f(x)\cdot{g(x)}=0\Rightarrow{f(x)=0\text{ or }g(x)=0}$ and remember that $e^{f(x)}\ne{0}$

c. You must find what values of x $v(t)>0$

d. You must find $\int_0^8\left|v(t)\right|dt$

e. You must find $a(3)=v'(3)$

f. You must find all values of x such that $a(t)>0$ and $a(t)<0$