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Dynamics Question using nonconstant acceleration 
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#1
Feb1910, 09:52 AM

P: 3

1. The problem statement, all variables and given/known data
A particle has an initial speed of 27 m/s. If it experiences a deceleration of a = (6t) m/s^2, where t is in seconds, determine the distance traveled before it stops. 2. Relevant equations a = dv/dt v = ds/dt ads = vdv (not independent from the above two equations) 3. The attempt at a solution What I know: v(0) = initial speed = 27 m/s v = final speed = 0 m/s t(0) = initial time = 0 seconds (assumption) s(0) = initial displacement = 0 meters (assumption) I used a = dv/dt and integrated to find the time for the particle to stop. I found it this way: a = dv/dt = (6t). dv = (6t)dt. Lower limit for v(0) = 27, upper limit for v = 0. Lower limit for t(0) = 0, upper limit for t = t. Integrating dv = (6t)dt, I get v(0) = 3t^2 > t = 3 seconds. This is the time it takes the particle to stop. However it asks for the distance to stop, and I have no idea how to get it. I know I can't use constant acceleration formulas because the acceleration is a function of time. I tried using ads = vdv as that is independent of time, but the acceleration a is dependent on time as a = (6t) m/s^2 


#2
Feb1910, 03:37 PM

P: 10

In your problem statement you state a constant decelaration of 6 m/s^2
Then you go on with: a = dv/dt = (6t), which is the correct defintion Integration produces : v = 3*t^2+v0. You got the integration idea very right. t = 3 sec. So, continue with: v = ds/dt in a similar way. v = ds/dt integration gives: s = (3/3)*t^3+v0*t+s0 You can solve this for t= 3 sec. 


#3
Feb2210, 05:39 AM

P: 3

Thanks man.
I'm 'relearning' dynamics for my FIT exam in October. Funny how the simplest solution is right there in front of you and yet, nothing. :) 


#4
Feb2210, 08:55 AM

P: 11

Dynamics Question using nonconstant acceleration
Why din you try 2nd Order Differential Equation? differentiate twice will induce many unnecessary constants.



#5
Feb2210, 09:32 AM

P: 3




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