Lamborghini Speed test Calculations

[QuickSkope]

Homework Statement

A lamborghini's breaking speed is 6.7 times that of its acceleration. The total time taken to go from 0 to 100 and back to 0 is 13.7 seconds. What is the displacement of the car?

Homework Equations

Unsure at the moment

The Attempt at a Solution

So, you divide the treck into 2 parts: Acceleration and Breaking

Acceleration:
a= x
t= y
d= ?
Vo = 0
Vf = 100

Breaking:
a= -6.7x
t= 13.7 - y (y being the accelerating time)
Vo= 100
Vf = 0
d=?

Honestly I am unsure as to what to do :S, any help would be much appriciated.

 

[SammyS]

Homework Statement

A Lamborghini's breaking speed is 6.7 times that of its acceleration. The total time taken to go from 0 to 100 and back to 0 is 13.7 seconds. What is the displacement of the car?

Homework Equations

Unsure at the moment

The Attempt at a Solution

So, you divide the trek into 2 parts: Acceleration and [STRIKE]Breaking[/STRIKE]

Acceleration:
a= x
t= y
d= ?
Vo = 0
Vf = 100

Breaking:
a= -6.7x
t= 13.7 - y (y being the accelerating time)
Vo= 100
Vf = 0
d=?

Honestly I'm unsure as to what to do :S, any help would be much appreciated.

Is this the wording given to you ?

"Braking speed" ? "Braking speed" ? "Braking speed" ? "Braking speed" ?

(I'll get off of that soap-box!)

First find the forward acceleration.

How are acceleration, change in velocity and time all related ?

 

[Nugatory]

With what you have already, you can calculate how much time the car spent accelerating from zero to one hundred and decelerating from one hundred to zero.

These times, and the change in speed which you already know, will give you the acceleration and deceleration.

Finally, there's an equation that gives you the distance covered in a given time at a given acceleration/deceleration, and that will take you to the answer.

As a historical note: 427 Cobras were doing zero to one hundred to zero in under ten seconds, more than forty years ago. Of course you would have to be totally completely barking mad, and suicidal as well, to push one of those cars to those limits... No ABS, no launch control, no traction control...

 

[QuickSkope]

Is this the wording given to you ?

"Braking speed" ? "Braking speed" ? "Braking speed" ? "Braking speed" ?

(I'll get off of that soap-box!)

First find the forward acceleration.

How are acceleration, change in velocity and time all related ?

Ya sorry, its braking acceleration :P. Anyway, Acceleration, Velocity and Time are related through the 4 big kinematic formulas:

V= Vo+ at
D = 1/2(v+Vo) * t
d = Vo*t + 1/2at^2
V^2 = Vo^2 + 2ad

Correct?

This is what I've come up with, might be absolutely wrong :S, sorry, JUST started physics:

1a + 6.7a = 13.7 sec
7.7a = 13.7 sec
a = 1.78

1.78 * 6.7 = 11.92 sec

11.92 + 1.78 = 13.7 sec

So it takes 11.92 seconds to accelerate, then 1.78 to get back to 0.

Then, using d = 1/2(V+Vo) * t

100 mph = 44.704 m/s

d = 1/2(44.704 + 0) * 11.92
d= 22.352 * 11.92
d= 266.43 m

Using it again:

d= 1/2(0 + 44.704) * 1.78
d= 22.352 * 1.78
d=39.78 m

266.43 m + 39.78 m = 306.22 m

Did I do it right or did I botch the entire thing?

EDIT: Nugatory, that's what I've done here (or tried to do :P). And Cobras are beauty cars, my friends dad actually has a Shelby Cobra, such an amazing car :)

 

[asteriatic]

As a scientist, my first step in solving this problem would be to gather all the necessary information and identify any missing variables. From the given information, we know that the breaking speed of the Lamborghini is 6.7 times its acceleration and the total time taken for the car to go from 0 to 100 and back to 0 is 13.7 seconds. However, we do not know the values for acceleration or displacement.

To solve for the displacement, we can use the equation d = v0t + 1/2at^2, where d is the displacement, v0 is the initial velocity, t is the time, and a is the acceleration. Since we are dividing the trek into two parts, we can solve for the displacement in each part separately.

For the acceleration part, we know that v0 = 0 and vf = 100, and we are missing the value for acceleration. However, we do know that the time taken for acceleration is y seconds. Plugging in these values into the equation, we get:

100 = 0 + (1/2)a(y)^2
100 = (1/2)ay^2
200 = ay^2
a = 200/y^2

Now, for the breaking part, we know that v0 = 100 and vf = 0, and we are missing the value for displacement. Since the time taken for breaking is 13.7 - y seconds, we can plug these values into the equation and solve for displacement:

0 = 100(13.7 - y) + (1/2)(-6.7x)(13.7 - y)^2
0 = 1370 - 100y - 68.3x(13.7 - y)^2
100y = 1370 - 68.3x(13.7 - y)^2
y = (1370 - 68.3x(13.7 - y)^2)/100

Now, we can substitute this value for y into our equation for acceleration to solve for the displacement in the breaking part:

d = 100(13.7 - y) + (1/2)(-6.7x)(13.7 - y)^2
d = 1370 - (1370 - 68.3x(13.7 - y)^2) + (-3.35x)(13.7 -