How to find the maximum velocity and maximum acceleration?

AI Thread Summary
To find the maximum positive coordinate, velocity, and acceleration of a particle described by the position function x=12t^2-2t^3, the first derivative for velocity, V=24t-6t^2, is set to zero to determine critical points. Substituting these critical points back into the position function yields the maximum positive coordinate. For maximum velocity, the same process is applied to the velocity function, while the maximum acceleration is found by taking the derivative of the acceleration function and setting it to zero. The fourth derivative of position, known as jerk, is also mentioned as part of the discussion. Understanding these derivatives is crucial for solving the problem effectively.
dumpman
Messages
17
Reaction score
0

Homework Statement


The position of a particle moving along an x-axis is given by x=12t^2-2t^3, what is the maximum positive coordinate, velocity and acceleration reached by the particle?

Homework Equations





The Attempt at a Solution


I took derivative of the quadratic and got V=24t-6t^2 and set it =0, then i solve for t and substituted in x=12t^2-2t^3 to get the max. positive coordinate.
But I don't know what to do to solve for the max velocity and acceleration:confused:
 
Physics news on Phys.org
You have the velocity as a function of time, just like you had the position as a function of time.
 
more hints?
 
Why not repeat the same process for your function of velocity?
 
Oh, so i set a=0 to find max velocity, but then what do I do to find the max acceleration?
 
Find the derivative of acceleration as a function of time and set it to zero. The physical term for the fourth derivative time is called jerk.
 
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