Velocity & Definite Integral

I am not sure how to approach this problem.

**Problem**

Suppose a certain object moves in a straight line with the following velocity, where is in meters per second and is in seconds:

Without using your calculator, but instead using properties of definite integrals and facts you know about area, determine the net change in distance of the object from to time and find the object's average velocity on this interval.

My impression is that the net change in distance would be:

But since we cannot use our calculators I am unsure.

Re: Velocity & Definite Integral

Quote:

Originally Posted by

**Biff** My impression is that the net change in distance would be:

But since we cannot use our calculators I am unsure.

Why are you taking the absolute value of v(t)?

Re: Velocity & Definite Integral

Would the net change in distance simply be displacement?

Re: Velocity & Definite Integral

Quote:

Originally Posted by

**Biff** Would the net change in distance simply be displacement?

I think so. "Net" means "ultimate," "final."

Re: Velocity & Definite Integral

Quote:

Originally Posted by

**Biff** I am not sure how to approach this problem.

**Problem**

Suppose a certain object moves in a straight line with the following velocity, where

is in meters per second and

is in seconds:

Without using your calculator, but instead using properties of definite integrals and facts you know about area, determine the net change in distance of the object from

to time

and find the object's average velocity on this interval.

My impression is that the net change in distance would be:

But since we cannot use our calculators I am unsure.

Integrate v to get distance. Integral is net area.

Distance = net area under line y = -2 + t from 0 to 6, net area of sine term is zero.

Avg velocity = net area from above/6

EDIT: That was net distance from starting point. TOTAL distance travelled is integral | v |, or total area, which would include the sine term.

2 Attachment(s)

Re: Velocity & Definite Integral

I hate to disagree with all y'all, but Biff's formula is correct. What the rest of you are calculating is the *displacement* of the object. Displacement also contains the direction. A quick example. If you go around a circle the displacement is 0, but the distance is the circumference of the circle.

What I would do with this is to graph the d(t). Look for the points on the graph the lie below the t axis. Make these positive. I have attached two graph to give you a better idea. The first is just a parabola. The second is abs(parabola). This is the one you want to use to find the distance.

-Dan

Re: Velocity & Definite Integral

The particle is travelling in a straight line, so you can treat velocity as a scalar.

They are looking for NET distance travelled:

So if my grocery store is two miles down a straight road and I go to get my groceries, my NET distance is 0 (Integralvdt) and my TOTAL distance is 4 (IntegralIvIdt). If on the way I go backwards and forwards a couple of times (the sine term), that adds to my TOTAL distance but leaves my net distance unchanged. Post 1 is asking for NET distance.

EDIT: You are given the graph of velocity vs time and Net distancE is Integvdt. The question does not ask for distance vs time. The question asks for NET distance travelled from 0 to 6, exactly as was done in my last post..

1 Attachment(s)

Re: Velocity & Definite Integral

Quote:

Originally Posted by

**Hartlw** The particle is travelling in a straight line, so you can treat velocity as a scalar.

My bad. The graph was simply for a demonstration, but it would have been more effective if I referred it correctly. Thanks for pointing that out.

As to the problem with the vectors...Unfortunately no. In one dimension vectors are often regarded as scalars. It makes the Math a bit easier and I've done it myself. But it really is a bad habit.

Displacement is the directed distance from point A to B. It is a vector. Distance is the length of the path traveled. Looking at the graph posted I've got "velocity" for the y axis, and naturally time for the x. Can we agree that the distance traveled must be either 0 or a positive number? Then on the graph what is the distance traveled from the times 0 to about 1.5 (seconds, I presume)? The "distance" you get in this case is *negative*. That makes no sense at all since distance must be positive. So it has to be a displacement. What the graph is telling us is that even though we (Physicists) can often treat vectors as scalars we really need to be more careful in our applications. (Especially when we're teaching it!)

Why the absolute value in the formula I gave? Whenever the displacement in the graph goes to the negative then the particle has reversed its direction. That has to somehow add to the distance traveled and is why we need to use the absolute value of the velocity...to make up for that difference.

-Dan

Re: Velocity & Definite Integral

Quote:

Originally Posted by

**Biff** I am not sure how to approach this problem.

**Problem**

Suppose a certain object moves in a straight line with the following velocity, where

is in meters per second and

is in seconds:

Without using your calculator, but instead using properties of definite integrals and facts you know about area, determine the net change in distance of the object from

to time

and find the object's average velocity on this interval.

My impression is that the net change in distance would be:

But since we cannot use our calculators I am unsure.

Rectilinear motion has only one component and physics or mechanics texts treat it as a scalar, ie, d and v rather than **d** and **v**; **v** = v**i** and then deal with v, and | v | means absolute value.

If I head east for 1 hr at 50mph (pos v) and then turn around and head west at 40 mph (neg v) for ½ hour, how far am I away from home (net distance)? How many miles have I put on my car?

Plot v vs time: ^{v} ¯¯¯¯|__ t . Integral vdt is net area (distance away from home) and Integral |v|dt is total area (distance, speedometer reading) after time t. Distance from home (Net distance) = 1x50 – 1/2x40 = 30miles. Speedometer increase (Total Distance) = 1x50 + 1/2x40 = 70miles.

That is why sine term drops out of post 1 (which asks for NET distance), leaving only the NET area under 2-t from 0 to 6 (one positive area and one negative area: ½ (2x2) – ½(4x4) = -6. For total area (distance), use |v| which now includes the sine terms.

Reference: Rectilinear motion in any physics or calculus text.

Re: Velocity & Definite Integral

In general, assume you start at origin and move along a curve going through origin. A positive direction for distance s is defined along curve starting from origin.

1) I_{0}^{t} **v** dt = I_{0}^{t} (d**r**/dt) dt = **r**, position at time t

2) I_{0}^{t} |**v**| dt = I_{0}^{t} (|ds|/dt) dt = |s|, total distance travelled along curve at time t, positive.

3) I_{0}^{t} (ds/dt) dt = s, distance from origin along curve at time t, positive or negative.

I stands for integral.

**r** = x(t)**i**+y(t)**j**+z(t)**k**

**v** = d**r**/dt

|ds| = sqrt(dx^2+dy^2+dz^2) = |**v**|dt

ds = +- sqrt(dx^2+dy^2+dz^2) = +-|**v**dt|, + or - depending on whether v points along positive or negative direction along curve.

If you start from origin and always move in a positive direction along curve there is no difference between 2) and 3). On the other hand, if you move in positive direction from 0 to t1, then negative direction from t1 to t2, etc, there will be a difference. 2) will give total distance travelled along curve and 3) will give distance from origin along curve, which could be positive or negative.

Re: Velocity & Definite Integral

Hartlw: I am not going to do this with you again. The OP's question has been answered and I'm not going to tolerate you (or anyone else for that matter) hijacking it. If you think I'm wrong then start a new thread.

-Dan

Re: Velocity & Definite Integral

Quote:

Originally Posted by

**topsquark** Hartlw: I am not going to do this with you again. The OP's question has been answered and I'm not going to tolerate you (or anyone else for that matter) hijacking it. If you think I'm wrong then start a new thread.

-Dan

Who answered it and what was their answer? If you are referring to your answer, it was wrong. We can discuss it in a new post if you wish.

I was the only one to answer the OP explicitly, directly, and correctly. In my last post I addressed the general question lucidly and precisely, which should be interesting and educational and for which you should be grateful.