college calc question need help?

Question

There is a biker riding constant velocity at k m/s the distance is traveled in t seconds given by the function d(t)=kt. There is a support car that starts when it passed and catches up so when it does it is traveling at the same speed its function is given by s(t)=6t^2-t^3. When the car catches up it immediately slows down. I've figured k=8m/s and t=2sec.

Suppose the biker is riding at a constant velocity (v) which can be different then k.

What is the expression that gives the time or times the car and the bike meet in terms of v?

Answer 1

I'm not sure I understand the question.
If the biker rides at 8 m/s the car catches him in 2 seconds, that's true, but the car speed at t=2 is 12 m/s, which is faster than the cyclist.

If k=9 m/s, the car and the biker intercept at t=3 where the car has speed 9 m/s.

as long as the
d(t)=v*t where v is constant, they intercept at v=9 and t=3

if you allow v to be constant and treated as an unknown
the biker has equations of motion as
d(t)=v*t and vb(t)=v


the car has
s(t)=6*t^2-t^3
vs(t)=12*t-3*t^2

they will meet whenever
s(t)=d(t)
or
v*t=6*t^2-t^3
note that t=0 is a solution and a boundary condition as stated in the problem

the other two roots are related as
0=t^2-6*t-v
the times of intercept for t>0 are
t=(6+/-sqrt(36-4*v))/2
note that v<=9 If the biker is faster, the car will not intercept

the time they meet when vb(t)=vc(t) is related as

12*t-3*t^2=v
3*t^2-12*t+v=0
the roots are
t={12+/-sqrt(144-12*v)}/6
the larger root is when they have both the same displacement and same speed
in this equation v<=12, which is larger than 9 so the displacement equation sets the limit on v.

j