Time to decelerate over a given distance

[asp55]

Alright, so the problem I'm working on is basically as such:

An object with an initial Velocity of 95m/s slides horizontally, constantly losing velocity. After traveling 8 km it comes to a stop.

How long did the object take to travel 8 km.

So here's how I went at it:

δx = v₀ + .5at^2

δx+v₀ = .5(δv/t)t^2

2(δx+v₀) = v(t^2)/t

2(δx+v₀)/δv = (t^2)/t

2(δx+v₀)/δv = t

Then I plugged in:
δx = 8000m
v₀ = 95m/s
δv = -95m/s

Which returned t = -170.421s (Which I thought was odd to begin with, as how can one have a negative value for time.)

But I tried plugging that into my initial equation to see that it all worked out but instead of getting δx = 8000m I got δx = 8190m

So obviously there is a flaw in my logic SOMEWHERE. I just don't know where.

Help please!?

 

[HallsofIvy]

You appear to be using a 'special character' that shows up on my reader as ? .
Was it δ ?

Your first error is not "logic" but the wrong formula: δ x= v₀t+ (1/2)at². You forgot the "t" multiplying v₀.
Making that change, assuming that "constantly losing velocity" mean that the decelleration is constant, δx= v₀t+ (1/2)at². You are given that x= 8000 m and that v₀= 95 m/s. That leaves a and t as unknowns. Since there are two unknowns, you need a second equation: in that same time, t, the velocity reduces from 95 to 0 so -95= at is the second equation.

You need to solve 8000= 95t+ (1/2)at² and -95= at.

Substitute t= -95/a as you did and solve the quadratic equation.

 

[asp55]

Ahh, so I did. Thank you.

 

[derekmohammed]

I used the formula

Vf^2 = Vo^2 +2ad
I found a from this. which was -0.564m/s^2

From there I went to the formula d= Vo+1/2at^2 and solved for time.
In which I got about 167 seconds.