Kinematics Issue: Solving Distance Traveled in 15s

[GT1981]

Hi all,

I have (what should be) a relative simple question regarding the use of Kinematic equations:

The problem is stated as:

A station wagon, initially traveling east at 30m/s, decelerates uniformly until coming to a complete stop at a traffic light. How far was the driver from the traffic light, if it took 15 seconds for the vehicle to come to rest after the driver applied the brakes?

I know the answer is 225m...but the solutions used this equation to solve it:

X-Xo = [ (Vo + V)/2 ]* t
= [(30m/s + 0m/s)/2] *15s
= 225m

This makes sense, but why can't one use this equation to solve it:
X- Xo = Vot + 1/2gt^2
= 30m/s*15s + 1/2*10m/s^2*(15s)^2 ??

HELP! Thanks!

 

[xAxis]

Where does 10m/s^2 comes from in your equation??

 

[GT1981]

Hi,

The 10m/s^2 is just the acceleration due to gravity (normally 9.8m/s^2)

 

[xAxis]

well, that's why you can't use this equation. We don't know what is the acceleration of the train and there is no reason to suppose it is equal to g.
You have used the equation for distance travelled, s = v0t + 1/2 at^2, but we have two unknowns, s and a, so we cannot solve this problem using that equation

 

[GT1981]

Oh, ok, so you can't assume g = 10m/s^2 unless explicitly stated then?

 

[xAxis]

I suppose you can't, but it has nothing to do with your question

 

[Integral]

You use g in problems involvng FALLING bodies. A car resting on the road is not falling. Its acceleration is independent of gravity.

You have a PM.

 

[GT1981]

A Pm??

 

[GT1981]

Better yet, why couldn't you just use DV = DX/ Dt ? That doesn't give you the correct answer though... if you use 30m/s = DV and 15s = Dt

 

[GT1981]

Ok, sorry about all those questions, I'm new to all this, but I suppose that DV = DX/Dt would give you the roundtrip value of 450. So if you halve that to find out just the distance to the stoplight, that would give you the expected answer of 225..
Thanks again!