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 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: 20,751 Thanks: 2135 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
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 Also, what did you mean by "when the car stops"?
I'm guessing the car is at rest for a short period of time, then continues on its path.

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: $\displaystyle s(t) = t^3-12t^2+36t$, 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: $\displaystyle 3t^2-24t+36$ b)Find Acceleration at 5 seconds, which I found to be $\displaystyle 6m/s^2$ by using the second derivative: $\displaystyle 6t-24$ 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
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 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.
$v(t) = 0$

$3t^2-24t+36 = 0$

$3(t^2-8t+12)=0$

$3(t-2)(t-6) = 0$

car is instantaneously at rest at t = 2 and t = 6 seconds

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