Expert: Paul Klarreich - 6/21/2011

Question
I have a G6 Ponitac (2009)    My question is would it be physically possible for a car to accelerate to 35 miles per hour in a distance of .16 miles    It takes 3 minutes to walk this distance per Map Quest  This would be starting from a dead stop at a stop sign attempting to reach another stop sign .16 miles down  the road

Answer
Questioner: Lucy Fitch
Country: Michigan, United States
Category: Advanced Math
Private: No
Subject: speed
Question: I have a G6 Ponitac (2009)    My question is would it be physically
possible for a car to accelerate to 35 miles per hour in a distance of .16
miles    It takes 3 minutes to walk this distance per Map Quest  This would
be starting from a dead stop at a stop sign attempting to reach another stop
sign .16 miles down  the road
------------------------------------------
Start by stripping out the irrelevant stuff:

I have a G6 Ponitac (2009)    My question is   << irrelevant

would it be physically possible for a car to accelerate to 35 miles per hour
in a distance of .16 miles  

It takes 3 minutes to walk this distance per Map Quest    << irrelevant

This would be starting from a dead stop

at a stop sign attempting to reach another stop sign .16 miles down the road   << irrelevant


So your question is:

would it be physically possible for a car to accelerate to 35 miles per hour
in a distance of .16 miles starting from a dead stop?

The equations are:

s = 1/2 a t^2 + v0 t + s0
v = at + v0
and you have:  

v0 = 0 << dead stop, you said.
s0 = 0 << assume the first stop sign is s = 0.

So it reduces to:

s = 1/2 a t^2
v = at

And you want to solve these two equations for a and t.  You have, also:

s = 0.16, your distance to the second stop sign.
v = 35 ,  your velocity at that ................

0.16 = 1/2 a t^2
35 = a t

From these, derive a value of a.  Then compare that with the published (no, I don't know where) acceleration figures for your Ponitac.  (Ponitac?)