I first have to know whether what I did (calculation leading to your velocity's derivation) is correct or not. So, is it?


\(\frac{[2\frac{m}{s}(t^{2}-2th-h^{2})-6\frac{m}{s}(t+h)+5m]-[2mt^{2}-6mt+5m]}{h}

=\frac{[2\frac{m}{s}t^{2}-4\frac{m}{s}th-2\frac{m}{s}h^{2}-6\frac{m}{s}t-6\frac{m}{s}h+5m]-2\frac{m}{s}t^{2}+6\frac{m}{s}t+5m}{h}

=\frac{-4\frac{m}{s}th-2{m}{s}h^{2}-6{m}{s}h}{h}

=\frac{h(-4\frac{m}{s}t-2{m}{s}h-6{m}{s})}{h}

=-4\frac{m}{s}t-2{m}{s}h-6{m}{s}\)

I'm sure I have missed something but don't you mind pointing?

 

I first have to know whether what I did (calculation leading to your velocity's derivation) is correct or not. So, is it?

First, you've messed up the units from the getgo, which you didn't do in your prior post. So you do know how to do it, just need to be more careful.

\(\frac{[2\frac{m}{s}(t^{2}-2th-h^{2})-6\frac{m}{s}(t+h)+5m]-[2mt^{2}-6mt+5m]}{h}\)

Remember, t and h both have units of time. So your first term has units of distance (m) over time (s) multiplied by time-squared (t^2, th, or h^2) leaving you with distance-time (meter-seconds). You second term has units of distance over time multiplied by time leaving you with just distance (meter). Your third term has units of distance (meter). Your fourth term (the first term in the second set of square brackets) has units of distance multiplied by time squared (meter-second^2) while you second has units of distance-time and your third has units of just distance. Since all of these terms are added together, they MUST have compatible units. Since they don't, you KNOW the answer is wrong at this point and there is no point going further until this is corrected.

Second, how did you go from (t+h)^2 to (t^2 - 2th - h^2)?

It should be be

\(\frac{[2\frac{m}{s^2}(t^{2}+2th+h^{2})-6\frac{m}{s}(t+h)+5m]-[2\frac{m}{s^2}mt^{2}-6\frac{m}{s}t+5m]}{h}\)

Notice that each term in the numerator has units of distance.

In your second line the final term of "5m" should be subtracted, not added (this is in addition to the mistakes carried over from earlier). But then you treated it as though it were subtracted in going to the next line. So two mistakes that happen to cancel out -- lucky, but luck can't be counted on. You need to be much more careful (but no matter how careful you are, we all make these kinds of mistakes all too often).

\(=\frac{[2\frac{m}{s}t^{2}-4\frac{m}{s}th-2\frac{m}{s}h^{2}-6\frac{m}{s}t-6\frac{m}{s}h+5m]-2\frac{m}{s}t^{2}+6\frac{m}{s}t+5m}{h}
\)

Should be

\(=\frac{2\frac{m}{s^2}t^{2}+4\frac{m}{s^2}th+2\frac{m}{s^2}h^{2}-6\frac{m}{s}t-6\frac{m}{s}h+5m-2\frac{m}{s^2}t^{2}+6\frac{m}{s}t-5m}{h}
\)

Now combine like terms

\(=\frac{4\frac{m}{s^2}th+2\frac{m}{s^2}h^{2}-6\frac{m}{s}h}{h}
\)

and divide through by the denominator

\(=4\frac{m}{s^2}t-6\frac{m}{s}+2\frac{m}{s^2}h
\)

Notice that each term has units of distance/time.

In your third and subsequent lines you somehow brought the seconds units from the denominator up into the numerator.

\(
=\frac{-4\frac{m}{s}th-2{m}{s}h^{2}-6{m}{s}h}{h}

=\frac{h(-4\frac{m}{s}t-2{m}{s}h-6{m}{s})}{h}

=-4\frac{m}{s}t-2{m}{s}h-6{m}{s}\)

Notice how your terms have incompatible units. The first has units of distance while the second has units of distance-time^2 and the third has units of distance-time.

If the units don't work out, the answer is wrong. Period.

You are actually pretty close, just need more care in your work, that's all.

 

Now combine like terms

\(=\frac{4\frac{m}{s^2}th+2\frac{m}{s^2}h^{2}-6\frac{m}{s}h}{h}
\)

and divide through by the denominator

\(=4\frac{m}{s^2}t-6\frac{m}{s}+2\frac{m}{s^2}h
\)

Notice that each term has units of distance/time.

In your third and subsequent lines you somehow brought the seconds units from the denominator up into the numerator.



Notice how your terms have incompatible units. The first has units of distance while the second has units of distance-time^2 and the third has units of distance-time.

If the units don't work out, the answer is wrong. Period.

Please explain to me how did you divide

\(=\frac{4\frac{m}{s^2}th+2\frac{m}{s^2}h^{2}-6\frac{m}{s}h}{h}
\)

and come up with

\(=4\frac{m}{s^2}t-6\frac{m}{s}+2\frac{m}{s^2}h
\)

Incompatible units? So do you mean t and s are both the same unit (i.e. if what we plug in t, it should be in unit s (so for the t and s to be the same xD), e.g. 5 s ) what if I plug 1 hour? should i convert it to 3600 s first before i can plug it? thanks!

---------------------
EDIT: Sorry you just reaaranged the terms (sorry!!!,,,, why didn't I figured it out?)

 

If you plug in 1 hour then you still have units of distance/time, but now you have

4 (meter-hours) per (second-squared) for the first term

and

6 (meters) per (second)

Both are perfectly valid measures of velocity, but you can't combine them together. Just as you can't combine inches and feet directly even though both are perfectly valid measures of distance. To be added or subtracted, the units must not only be compatible, they must be identical.

To be multiplied or divided, the units don't have to be identical or even compatible. If they are compatible but not identical, then you end up with extra units hanging around (like hours/second in the first terms above). These can be easily removed by multiplying the terms by the appropriate value of 1.

\(
4 \frac{m \cdot h}{s^2} \cdot \frac{3600s}{1h} \ = \ 14400 \frac{m}{s}
\)

 

I just can't understand how to identify compatible and incompatible units?


Notice how your terms have incompatible units. The first has units of distance while the second has units of distance-time^2 and the third has units of distance-time.

what makes it incompatible? meaning each term should just have a one unit? I mean for example each term should be distance only, or should be velocity only?

 

The units "feet" and "kilometers" are compatible because both are measures of the same thing, namely distance. But they are not identical.

Terms with compatible units can't be added unless they are also identical. But, you can always convert compatible units so that they ARE identical and so that you can add them.

Terms with incompatible units can't be added, period. No amount of converting will make them comptabible because they simply aren't measures of the same thing. A term with units of "mph" and a term with units of "lightyears" are not compatible -- the first is a measure of velocity and the second is a measure of distance. But "microns" and "lightyears" are compatible because both are measures of distance.

It's not having "just one unit" that matters. Note that "velocity" isn't a single unit, but rather a specific combination of two units, distance and time, namely "distance per time" or "distance/time" or "distance divided by time" (all three are equivalent ways of saying the same thing).

To see if they are compatible, reduce the units to the quantities they measure. So "feet" and "miles" both reduce to "distance" while "hours" and "seconds" both reduce to "time". In addition to "distance" and "time" you have "current" (as in electrical current) and "mass". It's fine to use derived units, such as "velocity" or "acceleration", but be aware that if you do this that there can be subtle conversions that make two compatible terms seem incompatible on the surface -- for instance, seeing one term with "velocity" and another term with "acceleration-time". If you use fundamental quantities, both of these have units of "distance/time".

 

I want to revive this thread.

At time -5s, you are 85 meters. At time -6s, you are 113 meters. Can you tell me how 85 meters goes to 113 meters? (The velocity is only -26 m/s)

I can understand why acceleration is 4 m/s^2 because at t=-5s, the velocity is -26m/s and at t=-6s, the velocity is -30 m/s. So -30m/s-(-26m/s) is 4m/s^2.

I just can't appreciate what velocity and acceleration means. I can only understand velocity if it does not change (i.e. constant) But acceleration, no really no.

I can just compute the acceleration and the velocity, but there are something in my mind that asks so, what, why, how, and so does it matter? I want these to be answered. Thanks!

 

When you are in a car that is getting on the highway, does it make sense to you to look at the speedometer and say, "Right now we are going 20mph. Oh, now we are going 40mph. Oh, now we are going 60mph." ? If so, then you can understand the notion of velocity even when it is changing.

Velocity is merely a measure of the rate at which the position is changing at a given moment in time. As you are going slowly, but speeding up, on the highway onramp you velocity is low, 20mph, which means that IF you were to travel at the same velocity that you are at right then, that your position would change 20 miles for every hour that you did so.

Acceleration is merely a measure of the rate at which your velocity is changing at a given moment in time and is very analogous. If your acceleration is 5mph/second, it means that whatever your velocity is right now, say 20mph, then one second from now it will be 25mph and two seconds from now it will be 30mph. If you were to continue at that acceleration, then in a minute you would be traveling at a velocity of 320mph. But if you let up on the gas as soon as you reach 60mph (in 8 seconds), then your acceleration drops to 0 and your velocity stops changing. Similarly, if you slow down until your velocity is 0, then your position stops changing.

 

Okay can you explain to me how 85 meters goes to 113 meters if the velocity at time -5s is only -26 m/s? Or my calculation was wrong?

Thanks for your precious time and brilliant answers and insights.

 

Okay can you explain to me how 85 meters goes to 113 meters if the velocity at time -5s is only -26 m/s? Or my calculation was wrong?

Thanks for your precious time and brilliant answers and insights.

But in the time from t = -5 to t= -6s, velocity change gradually from -26m/s at t = 5s to -32m/s at t = -6s. It is not a constant as you thought.

 

Ah, the velocity also changes in between two whole number t. But how it come to 113 meters? I mean is there a mathematical explanation aside from assuming that it does change gradually?

Edit: I thought it changes only after A second and not every possible fraction of a second.

 

Ah, the velocity also changes in between two whole number t. But how it come to 113 meters? I mean is there a mathematical explanation aside from assuming that it does change gradually?

Edit: I thought it changes only after A second and not every possible fraction of a second.

As in previous posts, we have:

\(
f(t) = 2 \frac{m}{s^2} t^2 - 6 \frac{m}{s} t + 5m
\)

\(
v(t) = \frac{df(t)}{dt} = 4 \frac{m}{s^2} t - 6 \frac{m}{s}
\)

f(t=-5s) = 85m
v(t=-5s) = -26m/s

f(t=-6s) = 113m
v(t=-6s) = -32m/s

Now from:

\(
v(t) = \frac{df(t)}{dt} = 4 \frac{m}{s^2} t - 6 \frac{m}{s}
\)

Or:

\(df(t) = v(t)d(t)\)

\(\int_{-5}^{-6} f(t)dt = \int_{-5}^{-6} v(t)dt = \int_{-5}^{-6}{(4 \frac{m}{s^2} t - 6 \frac{m}{s})dt = (2 t^{2} \frac{m}{s^2}-6t\frac{m}{s}) \mid _{t = -5s}^{t = -6s} = 28m\)

\(f( t = -6s) - f(t=-5s) = 28m\)

Therefore, if f(t=-5s) = 85m then f(t=-6s) = f(t=-5s) + 28 = 85m + 28m = 113m

 

It is continuously changing. When you are in your car accelerating onto the highway, you don't go at one speed for one second and then your speed instantaneously changes to a different speed each second. Your speed is changing smoothly and continuously. If nothing else, consider that your car doesn't know anything about "seconds". There is nothing special about 2 seconds compared to 1.738472 seconds or 2.018728 seconds.

The derivative IS the mathematical explanation.

Given the position of an object at two different times, you can calculate the average velocity of the object during that interval by subtracting the starting position from the final position and dividing the difference by the amount of time it took to go between them, right?

So if we have a function that tells use the position as a function of time, say x(t), then we can calculate the average velocity between two moments in time as

v(t,t1) = [x(t1) - x(t)]/(t1-t)

right?

If we define Δt to be (t1-t), then we have t1=t+Δt.

v(t,Δt) = [x(t+Δt) - x(t)]/Δt

So pick Δt to be some conveniently small value, say Δt=0.1s, and that

x(t)=2t^2-6t+5

What is the average velocity at t = -6.0s, t = -5.5s, and t = -5.0s?

Now this formula only gives you the average velocity over a span of time, but we can make Δt as small as we want. If we ask what the result is in the limit that Δt goes to 0, then we have a reasonable definition of the velocity at a specific moment in time.

v(t) = v(t,0) = Lim(Δt=0) [x(t+Δt) - x(t)]/Δt

But this is the definition of the derivative of x(t) with respect to t:

v(t) = dx(t)/dt

 

As in previous posts, we have:

\(
f(t) = 2 \frac{m}{s^2} t^2 - 6 \frac{m}{s} t + 5m
\)

\(
v(t) = \frac{df(t)}{dt} = 4 \frac{m}{s^2} t - 6 \frac{m}{s}
\)

f(t=-5s) = 85m
v(t=-5s) = -26m/s

f(t=-6s) = 113m
v(t=-6s) = -32m/s

Now from:

\(
v(t) = \frac{df(t)}{dt} = 4 \frac{m}{s^2} t - 6 \frac{m}{s}
\)

Or:

\(df(t) = v(t)d(t)\)

\(\int_{-5}^{-6} f(t)dt = \int_{-5}^{-6} v(t)dt = \int_{-5}^{-6}{(4 \frac{m}{s^2} t - 6 \frac{m}{s})dt = (2 t^{2} \frac{m}{s^2}-6t\frac{m}{s}) \mid _{t = -5s}^{t = -6s} = 28m\)

\(f( t = -6s) - f(t=-5s) = 28m\)

Therefore, if f(t=-5s) = 85m then f(t=-6s) = f(t=-5s) + 28 = 85m + 28m = 113m

I mean thank you it works. But how come definite integrals answers this??

 

Like a derivative, an integral is a simple mathematical concept applied to small steps taken in the limit that the step size goes to zero.

If you got at a velocity V from an amount of time T, how far have you travelled? D=VT, right?

Now, if I want to know how far I have travelled between time T1 and time T2 but my velocity is changing during that time, I can break that time interval up into a whole bunch of small intervals, Δt, and calulate how far I travelled, Δd during each time interval based on the velocity I was travelling at the beginning of each interval. Thus

Δd = v(t)*Δt

If then have to sum up all of these intervals starting at t=T1 and ending at t=T2.

The number of intervals I have is basically N=(T2-T1)/Δt

D = Ʃ{n=0 to N-1} v(T1+n*Δt)*Δt

But notice that my limits are effectively going from T1 to T2. As I make Δt smaller and smaller, I move from a finite sum toward an infinite sum (in the limit that Δt goes to 0) and, in doing so, we simply change the notation a bit.

D = ∫{t=T1 to T2} v(t) dt

 

\(\int_{-5}^{-6} f(t)dt = \int_{-5}^{-6} v(t)dt = \int_{-5}^{-6}{(4 \frac{m}{s^2} t - 6 \frac{m}{s})dt = (2 t^{2} \frac{m}{s^2}-6t\frac{m}{s}) \mid _{t = -5s}^{t = -6s} = 28m\)

\(f( t = -6s) - f(t=-5s) = 28m\)

Therefore, if f(t=-5s) = 85m then f(t=-6s) = f(t=-5s) + 28 = 85m + 28m = 113m

Your integral is going from t=-5s to t=-6s, which would be the integral evaluated at t=-6s (the upper limit) minus the integral evaluated at t=-5s (the lower limit). You limits in the integral need to be swapped.

 

Your integral is going from t=-5s to t=-6s, which would be the integral evaluated at t=-5s (the upper limit) minus the integral evaluated at t=-6s (the lower limit). You limits in the integral need to be swapped.

I don't understand why it need to be swapped. I used this formula:

\(\int_a^b f(x)dx = F(b) - F(a) \)

where F is the indefinite integral for a continuous function f(x)
and therefore:

\(\int_{-5}^{-6} f(t)dt = F(t = -6s) - F(t = -5s)\)

 

I don't understand why it need to be swapped. I used this formula:

\(\int_a^b f(x)dx = F(b) - F(a) \)

where F is the indefinite integral for a continuous function f(x)
and therefore:

\(\int_{-5}^{-6} f(t)dt = F(t = -6s) - F(t = -5s)\)

I messed up the second part of my question, which makes it hard to see the point I'm getting at.

So here's the point: Why are you integrating backward in time? Don't you want to integrate from the earlier time, t=-6s, to the later time, t=-5s, to find out how far the object moved as that one second period elapsed?

 

I messed up the second part of my question, which makes it hard to see the point I'm getting at.

So here's the point: Why are you integrating backward in time? Don't you want to integrate from the earlier time, t=-6s, to the later time, t=-5s, to find out how far the object moved as that one second period elapsed?

Yes, it is backward in time. I think it would be better to do it otherwise but don't think that it is wrong.

If we ignore direction and are only interested in distance, then it is good.

 

But if you are then adding the result of the integral, the sign makes a world of difference.

But I just looked and, actually, for what you were doing you did it right.

You were apparently trying to find f(t=-6s) given the knowledge of f(t=-5s) and the velocity as a function of time. In that case, you DID want to integrate from t=-5s (your starting point) to t=-6s (your ending point).

I latched onto one thing and didn't look at the details as hard as I should have.