Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Moving Particle


Date: 07/14/99 at 13:39:03
From: Kirsten Peterson
Subject: Moving Particle Equation

In my College Calculus class we were given the following problem. I 
think I understand the logic behind it but the answers just don't seem 
right. Could you please help me? This is the problem:

Assume a particle moves on the x-axis according to the formula

     x = t^3-6t^2+9t+5

Find:
   a. the velocity when t = 3
   b. the acceleration when t = 4
   c. the times when the velocity is zero
   d. the place where the acceleration is zero
   e. the acceleration at each of the two moments where the 
      particle is motionless

This is what I did:

First I differentiated the equation

   dx/dt = 3t^2-12t+9

   a. velocity when t = 3

I then plugged in 3 for t

   dx/dt = 3(3)^2-12(3)+9 = 0

   b. I do not remember how to find the acceleration of a particle.
   c. logically, with how I did a, the answer to this problem would be 
      the given in a, because the answer for the velocity is 0 so the 
      answer to this equation is going to be t = 3. This is where my 
      logic doesn't make any sense. How can that be the right answer?  

I pretty much stopped there because I got confused. Please help me?  
Thank you!

Kirsten


Date: 07/14/99 at 16:32:56
From: Doctor Anthony
Subject: Re: Moving Particle Equation

>Assume a particle moves on the x-axis according to the formula
>     x=t^3-6t^2+9t+5
>Find:
>a. the velocity when t=3

   dx/dt = 3t^2 - 12t + 9    and putting t = 3

   dx/dt = 27 - 36 + 9 = 0   so the particle is stationary at t = 3.


>b. the acceleration when t = 4

   accel = d^2(x)/dt^2 = 6t - 12   and putting t = 4

   accel = 24 - 12 =  12 m/sec^2


>c. the times when the velocity is zero.

   We solve  3t^2 - 12t + 9 = 0

               t^2 - 4t + 3 = 0

                 (t-1)(t-3) = 0

So the particle is stationary at t = 1 and t = 3.

>d. the place where the acceleration is zero.

   6t - 12 = 0

so   accel = 0  when t = 2

and then x = t^3 - 6t^2 + 9t + 5
           = 8 - 24 + 18 + 5
           = 7 metres

>e. the acceleration at each of the two moments where the particle is 
>motionless.

   When t = 1   accel = 6t - 12 = -6 m/sec^2

   when t = 3   accel = 16 - 12 = +6 m/sec^2

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
High School Calculus

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/