Stopping distance with nonconstant deceleration

[m0286]

Hi
Another question
It says:
A car is traveling at 72km/hr. At a certain instant its brakes are applied to produce a nonconstant deceleration of s"(t)=-t(in m/s^2) How far does the car travel before coming to rest?

Now there's a lot of questions like this in the book, however they use a constant deceleration which is given along with the velocity.

but what would i use as my deceleration? I believe there probably is, but to me now it seems like there isn't enough information to answer the question. I am not asking for the solution because i know that's not what your here for, but could someone give me a hint at what I do with this nonconstant deceleration? Thanks

 

[Hootenanny]

HINT:

$$s(t) = \int\int s''(t) \; dt$$

and this page may be of some use http://hyperphysics.phy-astr.gsu.edu/hbase/avari.html

HINT Number Two: You can find the time taken to come to rest by setting the velocity equal to zero.

 

[m0286]

still confused?

Sorry.. this question is driving me crazy
I just can't seem to understand it.. I've been staring at that website, and nothings making sense to me. Do I need to figure out the time first? because by those equations that's whut its looking like... But I can't figure out how to do that either.. This questions is just confusing me soo much ... Is there anything else you could say to help me out? Thanks

 

[Hootenanny]

Yes, you need to determine the time first, you can do this by finding the integral;

$$v(t) = \int -t \;dt$$

What is v before the breaks are applied? What is v when the car is stopped?

Edit: There's no need to apologise, we're all here to learn

 

[m0286]

I still don't get it!

Ok this could be very wrong, I am just not sure how to do it any other way.
(im going to use $ for the integral sign)

v(t)=$-t dt

=-1/2 t^2

so v(t)=-1/2 t^2 ( and I've changed km/h to m/s)
20m/s=-1/2 t^2
t=sq root( 20/-1/2) t=sq root(40) t= 6.324s


Since s"(t)=-t in (m/s^2)
does that just mean that the deceleration is -6.324m/s^2?

t is $$-t dt so -1/3t^3
and v(t)=-1/3t^3
t=sq root (20/-1/3) t=sq root(60) t=7.75s

is v=d/t d=v*t = 20m/s*7.75s=154m

This definately is not right... I am soooo confused i have no idea what's going on here, please HELP!

 

[HallsofIvy]

Ok this could be very wrong, I am just not sure how to do it any other way.
(im going to use $ for the integral sign)

v(t)=$-t dt

=-1/2 t^2

Almost: v(t)= -(1/2)t²+ C for some constant C.
If you set t=0, you get v(0)= C so C is the initial velocity. Since your initial value was "A car is traveling at 72km/hr. At a certain instant its brakes are applied to produce a nonconstant deceleration of s"(t)=-t(in m/s^2) How far does the car travel before coming to rest?", C= 72 km/h = 72000 m/hr= 72/60= 1.2 m/min= 1.2/60= 0.02 m/s and
v(t)= -(1/2)t²+ 0.02.

Since v(t)= dx/dt= -(1/2)t²+ 0.02,
x(t) is the integral of that: x(t)= -(1/6)x³+ 0.02t+ C. Taking x(0)= 0 to be the initial position, x(0)= -(1/6)x²+ 0.02t.
Now, "How far does the car travel before coming to rest?"
When does the car come to rest? That is, what is t when
v(t)= -(1/2)t²+ 0.02= 0. Using that value of t, then find x(t)= -(1/6)x²+ 0.02t.