Dynamics Question using nonconstant acceleration

gatrhumpy

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

poso

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.

gatrhumpy

Thanks man.

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

darkdream

Why din you try 2nd Order Differential Equation? differentiate twice will induce many unnecessary constants.

gatrhumpy

Probably could have, but haven't reviewed DEs yet. :)