Next: Velocity of a falling Up: Velocity and Acceleration: Put Previous: Velocity and Acceleration: Put

Position and velocity of a car

Example   In driving in a straight line from New York to Boston, your position function given in miles from New York is described by the function:

\begin{displaymath}f(t) = \frac {5 t^3}{3} - 25 t^2 + 120 t \end{displaymath}

where t is the number of hours since the trip began. It takes you 8 hours to get to Boston.


a) Find your velocity at time t= 1/2 hrs.


b) Do you ever backtrack during the trip?

Solution to a): First, we find your velocity function at any time by differentiating your position function.

v(t) = f'(t) = 5 t2 - 50 t + 120.

Then, to find your velocity after half an hour, we just plug 1/2 into the velocity function.

\begin{displaymath}v(1/2) = f'(1/2) = 5 (1/2)^2 - 50(1/2) + 120 = 5/4 - 25 +120 = 96.25 \text{
m.p.h.}
\end{displaymath}

Hey, it sounds high but a lot of states have upped their speed limits.


Solution to b) (Do you backtrack?): Yes you do. It seems you left your favorite Mets baseball cap at the rest stop when you took it off to check your thinning hair in the mirror. Passing through Providence, you noticed the shiny spot on your head when you looked in the rear view mirror and realized you had forgotten it. Unfortunately, when you got back to the rest stop, you found it in the toilet, where it had been left by a Phillies fan.

How do we see that you backtracked? Well, if the velocity function is ever negative during the trip (that is, when $0 \leq t \leq 8$), then you must have been backtracking at that point. But notice that

v(t)= 5 (t2 - 10 t + 24) = 5(t-6) (t-4).

In particular, v(t) = 0 when t= 4 hrs and t=6 hrs. That would be when you realized you lost your hat and turned around, making your velocity go from positive to negative, and also when you turned around at the rest stop after you realized that you could flush the hat goodbye. So we would expect that in between those times, you were backtracking and your velocity was negative. Just to be sure, let's check what the velocity was at time t = 5. Then,

\begin{displaymath}v(5) = 5 (5)^2 - 50(5) + 120 = -5 \text{ m.p.h.} \end{displaymath}

Yup, it's negative. You definitely backtracked. You really wanted that hat. But you probably could have made better time if you weren't backing down the shoulder of the northbound lanes. Oh, well....



Next: Velocity of a falling Up: Velocity and Acceleration: Put Previous: Velocity and Acceleration: Put
Joel Hass
1999-05-26