(adsbygoogle = window.adsbygoogle || []).push({}); 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

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# Dynamics Question using nonconstant acceleration

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