Asked by Morgan
Rectilinear Motion:
*Need help with these three!*
Directions: The position function of a particle moving on a coordinate line is given by the following eq'ns, where s is in feet and t is in sec. Describe the motion of the particle for any time. Make a DIAGRAM of the motion with respect to its position at time t.
1.) s(t) =t^3 -6t^2
3.) s(t) = 1+6t -t^2
5.) s(t)=t^3 -9t^2 +24t
*Need help with these three!*
Directions: The position function of a particle moving on a coordinate line is given by the following eq'ns, where s is in feet and t is in sec. Describe the motion of the particle for any time. Make a DIAGRAM of the motion with respect to its position at time t.
1.) s(t) =t^3 -6t^2
3.) s(t) = 1+6t -t^2
5.) s(t)=t^3 -9t^2 +24t
Answers
Answered by
Steve
#1 is t^2(t-6)
ds/dt = 3t^2-12t = 3t(t-4)
Since s(t) it has a double root at t=0, it starts by moving down, then up again at t=4.
#2 is just a parabola, with s(0) = 1
since ds/dt > 0 at t=0, s increases until t=3, then decreases from there
#3 ds/dt = 3t^2-18t+24 = 3(t-2)(t-4)
so, s(0) = 0, and s increases while 0<t<2, decreases where 2<t<4, then increases again from there.
Go to any good online graphing site and you can see the graphs.
ds/dt = 3t^2-12t = 3t(t-4)
Since s(t) it has a double root at t=0, it starts by moving down, then up again at t=4.
#2 is just a parabola, with s(0) = 1
since ds/dt > 0 at t=0, s increases until t=3, then decreases from there
#3 ds/dt = 3t^2-18t+24 = 3(t-2)(t-4)
so, s(0) = 0, and s increases while 0<t<2, decreases where 2<t<4, then increases again from there.
Go to any good online graphing site and you can see the graphs.
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