How to Solve a Basic Particle Motion Problem?

purplemarmose
Messages
2
Reaction score
0

Homework Statement


Moves at G(f)=t^2-4t+3
a. Find displacement at 2 sec
b. Find average velocity at 4 sec
c. Find instantaneous velocity at t=4
d. Find acceleration at t= 4

Homework Equations


Instantaneous rate of change
Displacement

The Attempt at a Solution


f'= 2t-4 (So this would be the velocity then?)
f''=2 (Acceleration?)

How do I go about finding it at specific values? Just plug it in?
For example...
Find average velocity at 4 sec:
2(4)-4= 4
 
Physics news on Phys.org
Hi purplemarmose! Welcome to PF :biggrin:

This is a physics problem, it would be more relevant to post it in the Introductory Physics section.

Anyway, I'll take a stab at it.

What you've done is correct. The first derivative does indeed give you the instantaneous velocity(instantaneous is important!) with respect to time. The second derivative gives the instantaneous acceleration. Of course, you have to plug in the time values for respective answers :-p
 
Infinitum said:
Hi purplemarmose! Welcome to PF :biggrin:

This is a physics problem, it would be more relevant to post it in the Introductory Physics section.

Anyway, I'll take a stab at it.

What you've done is correct. The first derivative does indeed give you the instantaneous velocity(instantaneous is important!) with respect to time. The second derivative gives the instantaneous acceleration. Of course, you have to plug in the time values for respective answers :-p

Thanks for your greeting! This problem was actually given to me by my calculus teacher! However, I do know it that it is relevant to physics. Thanks so much!
 
It might be helpful to at least copy the problem correctly. Surely it does not say G(f) and then give a formula in t. Then later, you talk about "f(t)". And "Find average velocity at 4 sec" makes no sense- an average velocity has to be over a given interval of time. I suspect you mean the average velocity between t= 0 and t= 4 although it might be between 2 and 4 since you were also asked for the displacement at t= 2. The derivative of a position function is a instantaneous velocity at the given value of t. And acceleration is the second derivative. It also should be evaluated at a given value of t, but here, the acceleration function is a constant.

Do you know the definition of "average velocity" between two values of t?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top