How Do You Calculate the Maximum Speed of a Cart Given Its Velocity Equation?

[AngusBurger]

Hi

I'm electrically-biassed, and we had a maths exam that was for both electrical and mechanical students. The last question on the paper was separated into two; one for electrical, one for mechanical. I did the electrical question (all about mean and RMS of an instantaneous voltage,) obviously, but was intrigued by the mechanical.

I've tried working it out but don't quite understand it, so I was wondering if you good people could help?

The speed of a cart is v=3+10t-2t². Determine the maximum speed.

I think that v is related to t by the last part of the expression, 2t², but after that I was stumped. There was no mention of taking gravity or friction into account, so I don't know if that's a part of the question.

Could anyone enlighten me as to how this would be done? Thanks.

 

[micromass]

Hi AngusBurger!

At a certain time t, the velocity is

[tex]v(t)=3+10t-2t^2[/tex]

To find the maximal velocity, we need to find the maximum of the function. To do that, we differentiate and we see when it is zero:

[tex]v^\prime(t)=-2t+10[/tex]

A quick analysis shows that the derivative is 0 if t=5 and before t=5 it is positive and after t=5 it is positive.
So v is increasing before t=5 and decreasing afterwards. So the maximal speed occurs at t=5 and equals

[tex]v(5)=4+10.5-2.5^2=4[/tex]

Hope I didn't make any mistakes!

 

[AngusBurger]

I follow you up until the last equation, but I'm at a loss as to where you got those figures from.

Edit: Unless that first 4 is supposed to be 3? But even then, is there a mathematical process I am missing for the final 4?

 

[Mark44]

Hi AngusBurger!

At a certain time t, the velocity is

[tex]v(t)=3+10t-2t^2[/tex]

To find the maximal velocity, we need to find the maximum of the function. To do that, we differentiate and we see when it is zero:

[tex]v^\prime(t)=-2t+10[/tex]

A quick analysis shows that the derivative is 0 if t=5 and before t=5 it is positive and after t=5 it is positive.
So v is increasing before t=5 and decreasing afterwards. So the maximal speed occurs at t=5 and equals

[tex]v(5)=4+10.5-2.5^2=4[/tex]

Hope I didn't make any mistakes!

Read the equation above as 4 + 10[itex]\cdot[/itex]5 - 2[itex]\cdot[/itex]5² = 4. IOW, the dots indicate multiplication, not decimal points.

Also, I'm sure micromass meant to say after t = 5, the derivative is negative.

I follow you up until the last equation, but I'm at a loss as to where you got those figures from.

 

[micromass]

And of course, I didn't calculate the derivative correctly Shame on me...

The derivative is

[tex]v^\prime(t)=10-4t[/tex]

Thus the maximum occurs at

[tex]t=5/2[/tex]

and equals 15,5.

 

[micromass]

This post corrects all of my horrible mistakes:

Hi AngusBurger!

At a certain time t, the velocity is

[tex]v(t)=3+10t-2t^2[/tex]

To find the maximal velocity, we need to find the maximum of the function. To do that, we differentiate and we see when it is zero:

[tex]v^\prime(t)=-4t+10[/tex]

A quick analysis shows that the derivative is 0 if t=2.5 and before t=2.5 it is positive and after t=2.5 it is negative.
So v is increasing before t=2.5 and decreasing afterwards. So the maximal speed occurs at t=2.5 and equals

[tex]v(5)=3+10*(2.5)-2*(2.5)^2=15.5[/tex]

[STRIKE]Hope I didn't make any mistakes![/STRIKE] I did.

I graphed it this time, and it appears to be correct. I'll reread my posts better next time... Sorry for the confusion.

 

[AngusBurger]

I guessed they weren't decimals.

Hmm...this seems a lot more complicated than it looks. I must be missing something pretty fundamental to be this lost.

Edit: Oh, now I've seen your edit I was understanding it perfectly well. I just got confused again. Sorry.

Edit II: Sorry, should that v(5) not be v(2.5) now?

In any case, thank you. I find with the electrical side of calculus that I know how to do it, rather than what I'm doing, so it's good to see the other side of the fence and force yourself to actually think about the processes involved. I still prefer trigonometry though.

 

[Mark44]

I guessed they weren't decimals.

Hmm...this seems a lot more complicated than it looks. I must be missing something pretty fundamental to be this lost.

Edit: Oh, now I've seen your edit I was understanding it perfectly well. I just got confused again. Sorry.

Edit II: Sorry, should that v(5) not be v(2.5) now?

Yes.

In any case, thank you. I find with the electrical side of calculus that I know how to do it, rather than what I'm doing, so it's good to see the other side of the fence and force yourself to actually think about the processes involved. I still prefer trigonometry though.

 

[HallsofIvy]

Since the velocity function, v(t)= 3+10t-2t², is quadratic, you don't need Calculus to find its maximum value- just complete the square:
[tex]v(t)= 3+ 10t- 2t^2= 3- 2(t^2- 5t+ 25/4- 25/4)= 3- 2(t^2- 5t+ 25/4)+ 25/2[/tex]
[tex]v(t)= 31/2- 2(t- 5/2)^2[/tex]
If t= 5/2, then v(t)= 31/2. If t is any other value then [itex]-2(t- 5/2)^2[/itex] is negative so v(t) is less than 31/2.