Particle 1 is subjected to an acceleration a=-kv, particle 2 is subjected to a=-
ID: 2992689 • Letter: P
Question
Particle 1 is subjected to an acceleration a=-kv, particle 2 is subjected to a=-kt, and particle 3 is subjected to a=-ks. All three particles start at the origin s=0 with an initial velocity v0=10 m/s at time t=0, and the magnitude of k is 0.1 for all three particles (note that the units of k vary from case to case). Plot the position, velocity, and acceleration versus time for each paricle over the range 0<=t<=10 s.Explanation / Answer
a=v'=x'' (1) a=v'=-kv dv/v=-kdt ln(v)=-kt+C v(0)=v0 ln(v0)=C ln(v/v0)=-kt v(t)=v0*exp(-kt), and a(t)=-kv(t)=-k*v0*exp(-kt) s(t)=int(v(t)dt)= (-v0/k)*exp(-kt)+D , s(0)=0 0=(-v0/k)*1+D, D=v0/k s(t)=(v0/k)*(1-exp(-kt)) Then you can plot s(t), v(t), and a(t). Note that [k]=1/seconds, so the units do work out. a(t) is an exponential decay curve (goes to zero as t--> infinity) v(t) is an exponential decay curve (goes to zero as t--> infinity) s(t) is a limited growth curve (approaches a value, v0/k, as t-->infinity) (2) a(t)=-kt v(t)=int(a(t)dt)=(-k/2)*t^2+C v(0)=v0 v0=C v(t)=(-k/2)*t^2+v0 s(t)=int(v(t)dt)=(-k/6)*t^3+v0*t+D, s(0)=0 0=(-k/6)*0+v0*0+D s(t)=(-k/6)*t^3+v0*t [k]=meters/second^3 a(t) is linear decay v(t) is quadratic decay s(t) is cubic decay (3) a=s''=-ks which has the solution of a complex exponential, or sines and cosines. s(t)=A*cos(t*sqrt(k))+B*sin(t*sqrt(k)) s(0)=0=A*1+B*0, A=0 s(t)=B*sin(t*sqrt(k)), but that's not enough to way B is. v(t)=sqrt(k)*B*cos(t*sqrt(k)), v(0)=v0 v0=sqrt(k)*B, so B=v0/sqrt(k) s(t)=(v0/sqrt(k))*sin(t*sqrt(k)) v(t)=(v0)*cos(t*sqrt(k)) a(t)= (-v0*sqrt(k))*sin(t*sqrt(k)), which follows from integration or from a=-ks [k]=1/seconds^2 s(t) is periodic with a maximum value of v0/sqrt(k) v(t) is periodic with a maximum value of v0 a(t) is periodic with a maximum value of v0*sqrt(k)