Question

LAST velocity questions I promise! :). s(t) = 16t^3 + 12t^2-144t. t>0.Find the displacement (change in position) of the particle on the interval [ 0; 2 ]?
What is the overall distance traveled by the object on the interval [0,2]?
What is the furthest distance that the object has traveled away from its initial position on the interval [ 0, 2 ]?

Answer 1

ok displacement is easy, just\[d=s(t_2)-s(t_1)\]I got -64 for that, you should check me.
the furthest distance occurs when s'(t)=v(t)=0, which we already found was at t=3/2, so the maximum distance is\[d_{\max}=s(3/2)\] If you don't know why the max distance occurs when s'(t)=0 you probably have not yet taken a Calculus class, or at least not a good one.

Answer 2

oh you needed total distance too, one sec...

Answer 3

its -112 for the first one

Answer 4

overall distance is 158 m

Answer 5

yes, you are right -112 is the first. I knew that, just wanted to make you check.. XP (no, typo in the calculator really)
I'll check the other...

Answer 6

haha ok

Answer 7

yes 158 is the other
nice job, you don't need me after all! :)

Answer 8

haha no i do! with the last one!

Answer 9

you don't get why the max distance is at s'(t)=v(t)=0
?

Answer 10

whats the equation for finding overall distance traveled on that interval?

Answer 11

and the equation for furthest distance?

Answer 12

have you taken calculus and if so, do you understand how to find max/mins of a function?

Answer 13

get a function and factor it out and equal those to 0?

Answer 14

ive taken calculus last year and forget stuff (highschool) and now im taking calculus again in college and forget some stuff

Answer 15

to find the maxs/mins of a function f(x) we need to find when f'(x)=0
here we have s(t) so we find when s'(t)=v(t)=0
we already found that happens at t=3/2, so find
s(3/2)=-135
as far as a formula for total distance we have to take the absolute value of the displacement for each interval to avoid counting "negative" motion. The intervals again are
0<t<3/2 and 3/2<t<2, so\[\left| s(2)-s(3/2) \right|+\left| s(3/2)-s(0) \right|=23+135=158\]I'm not sure how to write a completely general formula, but the idea is that you must evaluate the displacement for each interval individually and tak the absolute value of each, then add 'em up.\[\left| d_{ab} \right|+\left| d_{bc} \right|+...=\sum d\] for every interval [a,b], [b,c], [c,d], etc.

Answer 16

oh ok i see. to find the furthest distance the object has traveled...do i use the maximum somehow?

Answer 17

Let's be careful about the terminology here. By "furthermost distance" I take it you mean maximum displacement (the farthest the particle ever was from it's initial point), so yes, that is a maximum. We got that maximum of our displacement function by doing what you may have forgotten from calculus, setting the derivative of that function equal to zero:
s(t) = 16t^3 + 12t^2-144t
is our displacement function
s'(t)=48t^2+24t-144=0=v(t)
is the derivative of that function. Solving this for t tells us when we are at either a maximum or minimum. To find whether it's a max or min requires more work usually, but here since we know the particle started at s(0)=0, it doesn't matter, this must be a max.

Answer 18

oh, don't forget to plug the value you get for t from
s'(t)=0
into s(t) and NOT s'(t), since that would give you zero again.

Not to confuse you but since s'(3/2)<0 and s''(t)>0 we are technically at a minimum for the graph of s(t), but since the question asks for the furthest distance the particle traveled, it doesn't matter whether it went in the negative direction or the positive one, it's still the "maximum displacement".
If that statement confused you ignore it, you can get by without that concept for now.

Answer 19

ohkay got it. thank you so much i really appreciate all the effort you put into this!

Answer 20

no prob, you seem like you'll do fine in class. good luck!