Particle Velocity and Acceleration Equations for Position Function s(t)

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Homework Help Overview

The discussion revolves around determining when the velocity of a particle, described by the position function s(t) = 4t^3 - 8t^2 + t - 50, is zero. The context involves concepts of calculus, specifically derivatives related to velocity and acceleration.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss deriving the velocity and acceleration functions from the position function. There is an exploration of using the quadratic formula to solve for when the velocity is zero, with some participants questioning the correctness of their calculations and simplifications.

Discussion Status

The discussion includes attempts to apply the quadratic formula to find the times when the velocity is zero. Some participants have provided guidance on simplifying expressions, and there is an ongoing exploration of different approaches to arrive at a solution.

Contextual Notes

There is mention of discrepancies between participants' results and a provided answer in a textbook, which raises questions about simplification and accuracy in calculations.

1irishman
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Homework Statement



When will the velocity be zero?
t is in minutes
s is in metres



Homework Equations



s = 4t^3 - 8t^2 + t - 50




The Attempt at a Solution



I came up with the velocity equation by taking the derivative of function s.
I came up with the acceleration equation by taking the derivative of the velocity function. I don't know how to figure out(solve) when the velocity of the particle will be zero? Help please?
The velocity at t minutes is given by:
v(t) = 12t^2 - 16t + 1
Acceleration is given by:
a(t) = 24t - 16
 
Last edited:
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1irishman said:
The velocity at t minutes is given by:
v(t) = 12t^2 - 16t + 1
Looks good. So plug in the value for v and solve the equation for t.
 
can i use quadratic formula, like this?

t = -b + - sqrt b^2 - 4ac/2a

where a = 12, b = -16, and c = 1 ?
 
Exactly, that will give you the times at which v(t) = 0
 
1irishman said:
can i use quadratic formula, like this?

t = -b + - sqrt b^2 - 4ac/2a

where a = 12, b = -16, and c = 1 ?
Yes!
 
My answer came to be:

t = 16 + - sqrt208/24

but the answer in the back of book is 4 + - sqrt13/6

I am not sure how they got that?
 
Try simplifying your square root expression more that you already have, and you'll find you have the same answer.
 
oh yes thank you i see 16*13 = 208
 

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