Question

s(t) = t^2 - 6t + 5 models the motion of a person cycling along Rte 66 where s(t) is the number of miles north of Los Angeles the person is at time t hours.

1) Write functions for the cyclist's velocity and acceleration at any time t.

2) Find the position and velocity of the cyclist after: (Be sure to interpret your findings in terms of the problem.)
a) one hour
b) two hours
c) three hours
d) five hours
e) eight hours

3) What is the acceleration of the cyclist after each time? Explain these results in terms of the problem.

Answer 1

since you are taking Calculus, you MUST know that velocity is the derivative of distance, and acceleration is the derivative of velocity

1. so take those derivatives

2. a) plug in t = 1 into both your distance and your velocity expression
etc

3. That should be a constant. After all if you differentiate a quadratic term twice you end up with a constant
i.e. t^2 ---> 2t ---> 2

Answer 2

s(t)=t²-6t+5 is the position vector.
For example, at time=0 hour,
s(0)=5, so the person is 5 miles north of LA, etc.

Velocity is the derivative of position with respect to time, namely,
v(t)=s'(t)=2t-6
At time t=0,
v(t)=-6, meaning he was cycling towards LA.

Acceleration is the derivative of velocity with respect to time, so
a(t)=v'(t)=2
At all times, acceleration is positive.

Will leave part (2) for you to practice on (interpretation).