Hi!!
Since it is not stated and also for easy understand of the problem we
will consider a Linear Motion with Constant Acceleration for the car;
this is a very good approximation of what is really happening.
Your question states that a car can go from 0 mph to 60 mph in 9.2
seconds. And you want to know the distance (D) that the car gone by
the time it reaches the final speed. To do that we can use the
following formula
D = (Vf + Vo)*t/2
where:
D is the unknown distance
Vf is the final speed
Vo is the initial speed
t is the elapsed time
In this case we have that:
Vf = 60 mph = 1/60 miles per second (mps)
Vo = 0 mph = 0 mps
D is unknown
t = 9.2 s
Then
D = (1/60)*9.2/2 = 0.0766 miles = 404.8 ft = 134.73 yd = 123.2 meters
For the second part of your question take into account the following formula:
a = ( Vf - Vo )/t = (1/60)/9.2 = 0.0018116 miles/s^2
So at this acceleration, when it reaches 57 mph (=19/1200 mps)) the
distance traveled will be:
d = (Vf^2 - Vo^2) / (2*a) =
= (19/1200)^2 / (2*0.0018116) =
= 0.0691914 miles = 365.33 ft = 121.77 yd = 111.35 meters
For a full set of equations that must be used in constant acceleration
motion refer to the following articles:
"Acceleration of a Car":
http://hypertextbook.com/facts/2001/MeredithBarricella.shtml
"Motion with Constant Acceleration":
http://webphysics.iupui.edu/152/152F06/152Basics/kinematics/kinematics.html
"Uniform Acceleration":
http://www.students.yorku.ca/~thejacob/acceleration.html
"Linear Motion with Constant Acceleration":
This one is great, you have nice explanations and it shows the
formulae derivation. You will find this very helpful.
http://math.umcrookston.edu/physics/1012/lessons/lesson3.pdf
Try also "AJ Constant Acceleration Motion Calculator", an online
program for calculating various equations related to constant
acceleration motion. Calculator includes solutions for initial and
final velocity, acceleration, displacement distance and time:
http://www.ajdesigner.com/constantacceleration/cavelocity.php
IMPORTANT NOTE:
The use of constant acceleration is an approximation, in the real
world the cars expends more distance than the theoretical that can be
found by this method (about 10% more).
So the actual distance for your case is about 135 meters (145 yd).
This difference is due some traction loss when you apply the power
(you need to break inertia and static friction at the beggining, etc).
At the following article you will find a plot that shows the 1996 BMW
318ti's acceleration. You will see that the acceleration has two
stages 0.23g (= 2.26 m/s^2) and 0.36g (=3.53 m/s^2), and the car
reaches 60 mph in 9.1 seconds. The theoretical acceleration (constant
acceleration aproximation) is 2.95 m/s^2 (=0.3g), that is a pretty
good approximation for the average acceleration showed at the plot:
"1996 BMW 318ti Performance":
http://www.randomuseless.info/318ti/performance.html
The following pages will give you some examples and background on this topic:
"Ask NRICH: Acceleration/distance/time problem":
http://nrich.maths.org/discus/messages/8577/7263.html?1071520520
"Formula One car - Performance - Wikipedia, the free encyclopedia":
http://en.wikipedia.org/wiki/Formula_One_cars#Performance
Search strategy:
car acceleration "0 to 100" distance
car acceleration "0 to 60" distance
constant acceleration motion formulas
I hope this helps you. Feel free to request for a clarification if you
find something unclear, I will be glad to give you further assistance
on this topic if you need it.
Best regards,
livioflores-ga |