How Do You Derive the Acceleration Function from a Given Velocity Equation?

AI Thread Summary
To derive the acceleration function from the given velocity equation v=√(2P/M)(√T), it is essential to recognize that P and M are constants while T represents time. The acceleration is defined as the derivative of velocity with respect to time. The correct differentiation leads to the acceleration function a=√(P/2MT). The original poster initially struggled with the derivation process but ultimately realized that treating P and M as constants simplified the calculation. The discussion highlights the importance of understanding variable roles in differentiation.
david98999
Messages
17
Reaction score
1

Homework Statement



Derive the function for the acceleration from this function

v=√(2P/M)(√T)

The answer is √(P/2MT)

I have tried many different attempts but I am still unable to reach this answer.
[/B]
I am sure the process to get to the answer is a simple one but for some reason I am unable to see it

Homework Equations


I use the normal method of deriving the function , for each variable i use anx^(n-1)

The Attempt at a Solution



We have 3 variables P , M and T

derivative of P

(1/2)P^(-1/2) =1/(2√P)

derivative of M

(-1/2)M^(-3/2)= -M^(-3/2)/2

Derivative of T

(1/2)T^(-1/2)= 1/(2√T)these answers do not make up the final sum as m is negative ,

I just need to understand the process to get to the answer for this particular sum , not the subject in generalI can derive and integrate functions (ex: 3x^2= 3(2)x^(2-1)=6x and 6x^2/2=3x^2)
its just that I do not understand how to derive this equation into the given answer ,
I have a mental block about this question.I would appreciate any help

 
Physics news on Phys.org
david98999 said:

Homework Statement



Derive the function for the acceleration from this function

v=√(2P/M)(√T)

The answer is √(P/2MT)

I have tried many different attempts but I am still unable to reach this answer.
[/B]
I am sure the process to get to the answer is a simple one but for some reason I am unable to see it

Homework Equations


I use the normal method of deriving the function , for each variable i use anx^(n-1)

The Attempt at a Solution



We have 3 variables P , M and T

derivative of P

(1/2)P^(-1/2) =1/(2√P)

derivative of M

(-1/2)M^(-3/2)= -M^(-3/2)/2

Derivative of T

(1/2)T^(-1/2)= 1/(2√T)these answers do not make up the final sum as m is negative ,

I just need to understand the process to get to the answer for this particular sum , not the subject in generalI can derive and integrate functions (ex: 3x^2= 3(2)x^(2-1)=6x and 6x^2/2=3x^2)
its just that I do not understand how to derive this equation into the given answer ,
I have a mental block about this question.I would appreciate any help

It's not clear what P and M stand for in the formula for velocity. Is T supposed to represent time?

The acceleration has a very specific definition: it is the derivative of the velocity with respect to time. If P or M are not functions of time, then they are treated as constants when taking any derivatives w.r.t. time.
 
SteamKing said:
It's not clear what P and M stand for in the formula for velocity. Is T supposed to represent time?

The acceleration has a very specific definition: it is the derivative of the velocity with respect to time. If P or M are not functions of time, then they are treated as constants when taking any derivatives w.r.t. time.

sorry the function is to find the velocity of an alfa romero car , p= the power in watts , m is the mass and t is the time , the question wants me to derive the velocity as a function of time
 
david98999 said:
the question wants me to derive the velocity as a function of time

So M and P are constants ... You should only differentiate with respect to time.
 
You are right I did as you said

√(2p/m)*√t

a=dv/dt=(2p/m)^(1/2) x t^1/2 = 2^(-1) x (2p/m) ^(1/2) x t^(-1/2)
= √(2p/(m(2)^2)t)= √(p/2mt)

I had a feeling it was a simple error , I'm sorry I wasted every ones' time.
Thank you for your help
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top