Finding acceleration in kinematics

Posted 9 years ago
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 How can I find acceleration if it is assumed constant? For example: a car slows down from 29m/s to rest in a distance of 71m.
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Posted 9 years ago
 Physics problems should be solved from first principles In[8]:= xsoln[t_] = DSolveValue[{x''[t] == -a, x'[0] == 29, x[0] == 0}, x[t], t] Out[8]= 1/2 (58 t - a t^2) In[10]:= Solve[{xsoln[tf] == 71, xsoln'[tf] == 0}, {a, tf}] Out[10]= {{a -> 841/142, tf -> 142/29}} In[11]:= N[%] Out[11]= {{a -> 5.92254, tf -> 4.89655}} 
Posted 9 years ago
 You can also go to Wolframalpha and enter reduce 29 = a*t & 71 = a/2*t^2 or feed Mathematica with r = Reduce[29 == a*t && 71 == a/2*t^2] N@r[[1]] N@r[[2]] This might be a more comprehensible way.
Posted 9 years ago
 you can use one of kinematic equations of motion "final velocity^2 = initial velocity^2 + 2 acc distance"  Clear[vf, vi, a, dist, t] eq = vf^2 == vi^2 + 2*a*dist; values = {vf -> 0, vi -> 29, dist -> 71}; (a = a /. First@Solve[eq /. values, a]) // N And Mathematica says  -5.92254 so it is about -6 m/s. It is negative since it is deceleration. Now you can also find the time, using the other kinematic equation "final velocity=initial velocity+acc*time" (you could also use "dist = (final vel+initial vel)/2 * time)" to solve for time) eq = vf == vi + a*t; Solve[eq /. values, t] // N And Mathematica says  {{t -> 4.89655}} So it took about 5 seconds to stop.