# Time to decelerate over a given distance

1. Jan 7, 2005

### 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:

&delta;x = v0 + .5at^2

&delta;x+v0 = .5(&delta;v/t)t^2

2(&delta;x+v0) = v(t^2)/t

2(&delta;x+v0)/&delta;v = (t^2)/t

2(&delta;x+v0)/&delta;v = t

Then I plugged in:
&delta;x = 8000m
v0 = 95m/s
&delta;v = -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 &delta;x = 8000m I got &delta;x = 8190m

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

Last edited: Jan 7, 2005
2. Jan 7, 2005

### HallsofIvy

Staff Emeritus
You appear to be using a 'special character' that shows up on my reader as ? .
Was it &delta; ?

Your first error is not "logic" but the wrong formula: &delta; 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, &delta;x= 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.

3. Jan 7, 2005

### asp55

Ahh, so I did. Thank you.

Last edited: Jan 7, 2005
4. Jan 7, 2005

### derekmohammed

I used the formula