Derive kinematics + motion w/ calculus
Given Position function \(s(t)\)
Deriving Velocity function
\[ s'(t) = \frac{ds}{dt} = v(t)\]
Deriving Acceleration function
\[ s''(t) = v'(t) = \frac{dv}{dt} = a(t)\]
1 1st equation of kinematics
Given constant acceleration and initial velocity, find the find final velocity
Notice how this doesnt work DSolve[{s''[t]==-9.81, s'[0]==0},s'[t], t]
but this does DSolve[{v'[t]==-9.81, v[0]==0},v[t], t]
\[\{\{v(t)\to -9.81 (1. t+0.)\}\}\]
2 2nd equation of kinematics
Given constant acceleration = 13, derive equation for position
DSolve[s''[t]==13, s[t], t] // TeXForm
\[\left\{\left\{s(t)\to \frac{13 t^2}{2}+c_2 t+c_1\right\}\right\}\]
3 3rd equation of kinematics
Solve[{a==((v[t]-v[0])/t),s==((1/2)(v[t]+v[0])*t)},{v[t],v[0]}] // TeXForm
\[\left\{\left\{v(t)\to \frac{a t}{2}+\frac{s}{t},v(0)\to -\frac{a t^2-2 s}{2 t}\right\}\right\}\]
Simplify[v[t]^2 - v[0]^2 /. Solve[{a==((v[t]-v[0])/t),s==((1/2)(v[t]+v[0])*t)},{v[t],v[0]}]] // TeXForm
\[{2š¯‘ˇš¯‘ }\]
4 Newtonā€™s Second Law
Solving the differential equation F=ma
Solving for 1) distance function and 2) velocity function, each wrt time
\[m \frac{ds^2}{d^2t} = f\]
DSolve[m*s''[t] == f, s[t], t] // TeXForm\[\left\{\left\{s(t)\to \frac{f}{m}\frac{1}{2}t^2+c_2 t+c_1\right\}\right\}\] looks familiar \(s = \frac{1}{2} at^2 + v_0t\)
\[m \frac{dv}{dt} = f\]
DSolve[m*v'[t] == f, v[t], t] // TeXForm\[\left\{\left\{v(t)\to \frac{f t}{m}+c_1\right\}\right\}\] looks familiar \(v_n=at+v_0\)
