Calculus Problem: Braking Distance for Initial Speed 4v

[Shinaolord]

Homework Statement

The braking distance for an automobile with initial speed v is d. What is the braking distance if the initial speed of the automobile is 4v? Assume the automobile experiences the same constant acceleration in each case.

Homework Equations

v=v₀ + at
Δd = ∫v dt = v₀t + at²/2

The Attempt at a Solution

So, I know that the speed 4v is going to replace v₀, yielding
∫4v dt = 4∫v dt = 4(v₀t + at²/2) +c, where c can be considered d=0.
This is where I need some assistance. I'm not entirely sure what the next step I should take to solve the equation.
Could it possibly be that we isolate v, and then compute? I'm really confused on this question.
By the way, this is from an exam from my college used several years ago, they are now used as practice exams.

EDIT: Could it possibly be as simple as 4d? I have severe doubts that this is definitely not the right answer, and I'm worried that I haven't' developed the intuition to solve these problems yet...With exams starting The week of the 15th

 

[Shinaolord]

Possibly, 4v= 1/2 at²?
[STRIKE]than √8v/a = t₁

compared to √2v/a= t₂

and then plug these into the kinematic equation for distance?

[/STRIKE]

I retract this statement.

 

[collinsmark]

You could actually solve this using the d = v₀t +(1/2)at², along with some other substitutions. But that's not the easiest way. As a matter of fact, that way wouldn't be very easy at all (although technically possible).

Why not start with a different kinematics equation for uniform acceleration. For this problem you are not interested in time, t, so you really don't need an equation that has t in it.

But can you find a kinematics equation for uniform acceleration that has velocity (initial and final), acceleration and distance in it?

 

[collinsmark]

By the way,

If you don't know which equation that I'm talking about, you could find it by starting with your
v = v₀ + at
equation, solve for t and substitute that into the
d = v₀t + (1/2)at²
equation. So it is possible with the equations that you already have.

You can alternatively derive the equation for which I speak by using the work-energy theorem.
The work done for a single, constant force (which implies uniform acceleration, since F = ma) over a distance, d, is W = Fd
The change in kinetic energy is (1/2)mv² - (1/2)mv₀².
(equate the two and cancel out the ms).

 

[Shinaolord]

My apologies, I fell asleep, and with your suggestions I would now start with
v_f² =v₀²+2aΔx

 

[Shinaolord]

I will chug the numbers tomorrow, I'm to tired and will most likely make a mistake.