March 5th, 2016, 08:02 PM  #1 
Member Joined: Jan 2016 From: Canada Posts: 40 Thanks: 0  Acceleration
In order to find a time (t) to find out when the car stopped momentarily do I use the first or second derivative to solve for t? Because when the car stops, both velocity and acceleration will be zero. It looks easier to use acceleration equation than using the quadratic equation for velocity to solve for t. I am just not sure which one will give the correct answer.
Last edited by atari; March 5th, 2016 at 08:06 PM. 
March 5th, 2016, 11:55 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,390 Thanks: 2015 
The first or second derivative of what? Also, what did you mean by "when the car stops"? A car is stopped (momentarily or otherwise) if and only if its velocity is zero. If it's momentarily stopped, its acceleration isn't necessarily zero. If its acceleration is zero, its velocity isn't necessarily zero. 
March 6th, 2016, 01:50 AM  #3  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,770 Thanks: 627 Math Focus: Yet to find out.  Quote:
Atari do you have the function of displacement, velocity or acceleration? You only need one of these functions to find the others. Then you can use your equations of motion to solve for t.  
March 6th, 2016, 06:40 AM  #4 
Member Joined: Jan 2016 From: Canada Posts: 40 Thanks: 0 
The function is: $\displaystyle s(t) = t^312t^2+36t$, where s is distance in metres and t is time in seconds for t [0,10] a) Find velocity at 3 second, which I found to be $\displaystyle 9m/s$ by using the first derivative: $\displaystyle 3t^224t+36$ b)Find Acceleration at 5 seconds, which I found to be $\displaystyle 6m/s^2$ by using the second derivative: $\displaystyle 6t24$ c) Then I am asked to find At what times within those 10 seconds will the car stop momentarily? And this is where I am not sure whether to use the first or second derivative to solve for t. 
March 6th, 2016, 08:59 AM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 2,847 Thanks: 1486  Quote:
$3t^224t+36 = 0$ $3(t^28t+12)=0$ $3(t2)(t6) = 0$ car is instantaneously at rest at t = 2 and t = 6 seconds  

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