A level Mechanics: Variable acceleration differentiation

Please can someone help me with this question? I haven’t been taught differentiation in Pure maths yet, however in Mechanics we are using it to find expressions for the velocity and acceleration of a particle when given the displacement. In lesson, our teacher has given us a “crash course” on differentiation but hasn’t taught us it in much detail at all since we’ll do that in Pure maths later on. Therefore, my question may be simple but I am really stuck on it.

The question is:
Find an expression for (i) the velocity and (ii) the acceleration of a particle given that the displacement is given by: x = 3t ⁴ - 2t ³ + 5 all over 2t. I simplified this to 3/2 t ³ - t ² + 5/2 t however I don’t know if this is correct.

Any help will be very much appreciated because I’ve been stuck on it for ages and my teacher isn’t really the sort of person who makes it easy asking questions.

Thank you so much and merry Christmas everyone!

Reply 1

0gg

You know that on a displacement-time graph, the slope at a given point is the velocity at that point. x is the displacement, and t is the time, so the graph of x against t is like a displacement-time graph, but you know the function that represents the line! That makes it easy to find the velocity at a given point because the slope at the given point is the derivative.

Your simplification is perfect, and from there, you can find the derivative with the 'power rule': multiply the coefficient by the power, then decrease the power for each term. e.g. for 2t^5, the derivative is 10t^4, for 6t, the derivative is just 6 and so on.

I hope that explains differentiating polynomials! Differentiate the expression for displacement to get the velocity, and then you can differentiate a second time to get an expression for acceleration.

Merry Christmas!

Reply 2

davros

Original post by Starlettt

Please can someone help me with this question? I haven’t been taught differentiation in Pure maths yet, however in Mechanics we are using it to find expressions for the velocity and acceleration of a particle when given the displacement. In lesson, our teacher has given us a “crash course” on differentiation but hasn’t taught us it in much detail at all since we’ll do that in Pure maths later on. Therefore, my question may be simple but I am really stuck on it.
The question is:
Find an expression for (i) the velocity and (ii) the acceleration of a particle given that the displacement is given by: x = 3t ⁴ - 2t ³ + 5 all over 2t. I simplified this to 3/2 t ³ - t ² + 5/2 t however I don’t know if this is correct.
Any help will be very much appreciated because I’ve been stuck on it for ages and my teacher isn’t really the sort of person who makes it easy asking questions.
Thank you so much and merry Christmas everyone!

Just to clarify, is your displacement (3t^4 - 2t^3 +5) / (2t), in which case your simplification looks good :smile:

EDIT : you do mean 5/(2t) at the end don't you? :smile:

Rather than someone repeat the rules here, I would suggest googling "differentiating polynomials" to get you started. Most examples will show you how to find dy/dx when y is a function of x, whereas here you must remember that x is a function of t so you are finding dx/dt (for example). That will show you how to tackle the 1st 2 terms in your expression.

Now, 5/(2t) isn't a polynomial, but it can be rewritten as (5/2)t^(-1) - in other words, it's a constant times a power of t. The rule for differentiating in this case is just the same one you use for t^2, t^3 etc.

Have a go and post if you get stuck!

(edited 2 months ago)

Reply 3

Starlettt

Original post by davros

Just to clarify, is your displacement (3t^4 - 2t^3 +5) / (2t), in which case your simplification looks good :smile:

Rather than someone repeat the rules here, I would suggest googling "differentiating polynomials" to get you started. Most examples will show you how to find dy/dx when y is a function of x, whereas here you must remember that x is a function of t so you are finding dx/dt (for example). That will show you how to tackle the 1st 2 terms in your expression.
Now, 5/(2t) isn't a polynomial, but it can be rewritten as (5/2)t^(-1) - in other words, it's a constant times a power of t. The rule for differentiating in this case is just the same one you use for t^2, t^3 etc.
Have a go and post if you get stuck!

Omg thank you so much! It was the last term I was stuck on and could not think for the life of me how to do it. However, after a bit of thinking, it makes sense that 5/2t could be written as 5/2 t^-1 which should help me solve it correctly. Thank you so much for your help, it is really appreciated 😁

Reply 4

davros

Original post by Starlettt

No problem.

How come you're onto variable acceleration (which requires calculus), but you haven't done basic differentiation yet?
Are you in Y12 and your Mechanics teacher is really getting ahead, or Y13 and your Pure is way behind?

Reply 5

Starlettt

Original post by davros

No problem.
How come you're onto variable acceleration (which requires calculus), but you haven't done basic differentiation yet?
Are you in Y12 and your Mechanics teacher is really getting ahead, or Y13 and your Pure is way behind?

I am in year 12 but because our school is so small, in our mechanics lesson we do it with the year 13 (whom there are only 3 of), meaning we started halfway through the textbook. In pure maths, however, it is only year 12 so we are working through the textbook in the correct order. It is a strange prospect however in our school I understand that it works better. Hope this makes sense! Have a good night