Just think of it in reverse time: Start from v=0 and accelerate so as to be a v0 at distance d. If accel is constant c there's only one variable to solve two equations v=c*t d=0.5*c*t^2 =0.5*v^2/c But all you need to do is accelerate fast enough so you reach v before you reach d and then coast. Or in your case the reverse of that. Brent Meeker On 2/7/2014 8:09 PM, David Wilson wrote:
This is a linear motion problem (actually a work problem).
At time 0, a particle is at distance 0 and known velocity v (d(0) = 0, v(0) = v). At known time t, the particle comes to a stop at known distance d (d(t) = d, v(t) = 0). Velocity decreases over the entire motion (a(x) <= 0 for 0 <= x <= t). The particle never reverse direction (v(x) >= 0 for 0 <= x <= t).
I don't think this is soluble with constant acceleration (a(t) = c), perhaps with acceleration linear in time?
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