Equation of motion and Calculus

[Jamie S]

Hi all,

I started a level 3 btech in mech engineering and today was my first physics class. All went well apart from the tricky question at the end of class.

I thought it was a good idea to ask the question "how much harder can this be from last year".
Turns out for me not being brilliant at calculus it can be much harder.

The question surrounded the formula s=ut+1/2at^2
I was asked to differentiate s with respect to t, and show displacement is a minimum when u= - at

Off course it was a disaster and i laughed it off has not being forced to answer the question at which point other pupils attempted it.

We never did have it explained fully but it has made me wonder how one would go about answering this using differentiation. I've used algebra and transposed the formula before but this is the first time I've seen it this way.

It all got a little bit more confusing when someone suggested that it can't be differentiated until its been integrated. Is there any truth in this?

Cheers

 

[Doc Al]

We never did have it explained fully but it has made me wonder how one would go about answering this using differentiation

Well, you are given displacement as a function of t. Can you take the derivative? (How do you find the max / min values of a function using calculus?)

It all got a little bit more confusing when someone suggested that it can't be differentiated until its been integrated. Is there any truth in this?

No idea what that is supposed to mean. Ignore it.

 

[Jamie S]

the only time I've taken min and max values was by using a quadratic equation

 

[Doc Al]

the only time I've taken min and max values was by using a quadratic equation

This will be easier.

 

[Jamie S]

using quadratic formula was suggested however it was insisted that differentiation be used.

im looking at my notes from mondays math and the handout i received with calculus rules.
isnt the derivative of a function normally dy/dx = ...
in this case would that be;
ds/dt = ut+1/2at^2

 

[Doc Al]

in this case would that be;
ds/dt = ut+1/2at^2

No, that right hand side is s, not ds/dt. You start with the function s, which is given, and take the derivative.

 

[Jamie S]

i don't quite follow

 

[SteamKing]

using quadratic formula was suggested however it was insisted that differentiation be used.

im looking at my notes from mondays math and the handout i received with calculus rules.
isnt the derivative of a function normally dy/dx = ...
in this case would that be;
ds/dt = ut+1/2at^2

Or, more to the point, if s = ut + (1/2)*a*t² and you take the derivative of s with respect to t to find ds/dt, you must do the same to the right hand side of the equation.

 

[Chestermiller]

Are you saying that you have not yet had calculus?

Chet

 

[Jamie S]

ive had one mathematics class up to now. very little on my level 2.

 

[Jamie S]

you no i am aware that i will need to improve drastically if I am to ever progress from this level. at the moment through we've not touched on any calculus though we have only just enrolled monday.

 

[Chestermiller]

Here's what I understand from what you said in your previous two posts: You never had calculus in any of your courses before, and today is only your second day in a calculus course. If that is the case, then you are not going to be able to solve this problem using calculus.

Chet

 

[Jamie S]

Chestermiller... I take your comments and while i may or may not have had enough time using calculus my initial intention was never to solve this equation. I did however ask about it because i was curious of how it would be solved using the calculus method. It isn't homework or course work, i at least hope so given I've just started the course so i didnt see were there would be hesitation to answer.

I thank all that have made the effort to comment and suggest solutions.

 

[Chestermiller]

Chestermiller... I take your comments and while i may or may not have had enough time using calculus my initial intention was never to solve this equation. I did however ask about it because i was curious of how it would be solved using the calculus method. It isn't homework or course work, i at least hope so given I've just started the course so i didnt see were there would be hesitation to answer.

I thank all that have made the effort to comment and suggest solutions.

Hi Jamie,

I'm proud of your aggressiveness for being so interested in seeing how calculus might be applied to problem, even before you have studied calculus. Bravo. My advice is to just be patient and, in almost no time (probably in just a few weeks), you will be be solving problems like this on your own. Best of luck in your calculus course.

Chet