How do i solve this problem of finding distance when object is at rest?

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wduk

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Hello

This is a homework question so i won't give the full question because i need to solve it myself, but am a bit confused how to go about it logically.

I have an object starting at rest at time t = 0. I'm given an equation for the velocity, and told to work out the distance the object travels until it comes to rest again.

So i integrate velocity V equation to get the displacement S equation. But then i am lost on how i would go about finding the distance for when the object is at rest again.

Was hoping some one could give some advice on where i go from here to solve the question.

Thanks :)
 
What do we know about the magnitude of an object's velocity when it is said to be at rest?

Also, note that there is a difference between displacement and distance traveled. If I walk a mile down the road, then turn around and come back to where I started, my final displacement is 0 miles, but the distance I traveled is 2 miles. This means you want to integrate the absolute value of the velocity function to get total distance traveled. :D
 
Well the velocity of an object at rest is zero so the magnitude is zero.

I did integrate the velocity equation, but when you say the absolute value i'm assuming you mean something more than just simply integrating ? I am bit confused what you meant. Can you give an example?
 
wduk said:
Well the velocity of an object at rest is zero so the magnitude is zero.

I did integrate the velocity equation, but when you say the absolute value i'm assuming you mean something more than just simply integrating ? I am bit confused what you meant. Can you give an example?
Click to expand...

Suppose we are told:

\(\displaystyle v(t)=t(t-1)(t-2)\)

And we are asked to find the total distance \(\displaystyle d\) traveled on \(\displaystyle 0\le t\le2\). So, we observe that on the following intervals we have:

\(\displaystyle (0,1)\) we have \(\displaystyle 0<v(t)\)

\(\displaystyle (1,2)\) we have \(\displaystyle v(t)<0\)

So, we need:

\(\displaystyle \displaystyle d=\int_0^1 v(t)\,dt+\int_1^2 -v(t)\,dt\)

Of course, we should always look for any symmetries we can employ to simplify the process of integration. For example, supppose we let:

\(\displaystyle u=t-1\implies du=dt\)

And we now have:

\(\displaystyle v(u)=(u+1)u(u-1)\)

Hence

\(\displaystyle \displaystyle d=\int_{-1}^0 v(u)\,du-\int_0^1 v(u)\,du\)

Since \(\displaystyle v\) is now an odd function, we may write:

\(\displaystyle \displaystyle d=-2\int_{-1}^0 v(u)\,du=2\int_0^1 v(u)\,du\)
 
Not sure i understand your method of approach at least not how i was taught it.

I am not given a time interval do any form of sums between definite integrals here. I am told to state the distance travelled by the object before it next comes to rest. And i am only given an equation for velocity v. And also told that at time t = 0 the object started at rest.
 
Did you solve the velocity function for zero? We know that t=0 is one solution; what is the smallest positive solution?
 
The object starts to move at t=0, and the "object travels until it comes to rest again" at t=sqrt(2).

In other words, the time interval starts at zero and ends at sqrt(2).
 
Are you saying it's as simple as plugging in sqrt(2) in to my integrated function and it will give me the distance? If so then i feel silly for over thinking this lol...
 
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