Question

So i found v(t)= te^(-t/4)*(1/t-1/4). Now it wants to kno where the particle is at rest. I kno that I have to set it to zero but how to i factor this to solve it? Any help?

Answer 1

is that (te^(-t/4)) ( 1/t - 1/4)

Answer 2

yes it is...

Answer 3

The way I would resolve this is by noting that
a*0 = 0
So, you can ignore the te^(-t/4) and just find when 1/t-1/4 = 0
which occurs when t=4
So, the velocity of the particle will be 0 after 4 seconds... if you need to find where the particle will be, you'll need to integrate v(t) to find r(t) and see what the displacement is.

Answer 4

what do you mean integrate? there are more things it wants to kno such as when it is slowing down and speeding up, where it is after 8 sec's blah blah.... you were right about the t=4. nice trick. forgot about a*0.... but you kno more than me for the moment lol...

Answer 5

I have v(t) and the position... what is r(t)? and where is it coming from?

Answer 6

well set te^(-4/t) = 0 and (1/t - 1/4) = 0. first one use logs and second one should be simple enough

Answer 7

Sorry for complicating things a bit for you twignamanda... I assumed this was coming from a calculus based physics course.
v(t) is standard notation for velocity
r(t) is standard notation for displacement and
r'(t) = v(t) so integral v(t) = r(t).... but you apparently don't need to know that yet :P

As for speeding up and slowing down, that's a bit more annoying, you'll need to look at the slope of the equation.
For t=8... you can just plug 8 in for t and evaluate it.

Answer 8

in other words you dont need to facotr anything like you normally would just set each term individually equal to zero

Answer 9

can you find the derivitave of the velocity? here it is defined as the acceleration. Sorry, this one problem is giving me h e double L!

Answer 10

Sorry for the late response, but yes
v'(t) = a(t)
In your case:
v(t)= te^(-t/4)*(1/t-1/4)
v'(t)=a(t)=(t/16-1/2)*e^(-t/4)