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Finding acceleration in kinematics

Posted 9 years ago

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.

POSTED BY: Aaron Phillips
3 Replies

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.

POSTED BY: Nasser M. Abbasi
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 BY: Simon Tyran

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 BY: Frank Kampas

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