Velocity vs Deceleration

[dballeng]

I am looking to write a 'braking' rule for our car race (box car) that will define the distance in which a car needs to stop for safety. If I put a deceleration of 15ft/sec/sec, the kids might get confused. Can I write the following "All vehicles must have a braking system which stops vehicle and driver at a maximum rate of 15 ft/sec" and it still be valid?

[PinkertonD]

Ummm, my thoughts, no.

Also I suspect you mean minimum rate of - whatever.

How will anyone know the rate of decel? Are you measuring it? Who has the DIY equipment to do that?

Personally, I would give something a little more tangible that is simple to measure. I'd suggest...
Minimum braking distance for combined vehicle and driver is 29' from a steady 20mph.
Those are just "play" figures. The builders can use a GPS/smartphone/etc to measure speed then apply the brakes, grab a tape measure and get the distance. A very basic hiking-GPS can be had for under 50-bucks.

No sophisticated measuring equipment needed.

[zeke]

How about showing them a plot of speed vs time which is a negative sloping straight line
(slope being -15).
Then explain how the area under that curve is the distance traversed before stoppage.

[PinkertonD]

@Zeke, that's an even better idea than a fixed speed/time stipulation.

[OMICSGroup]

System 1 is the car.
Initially it's moving at 30 km/h
You don't say anything about friction, so after the car separates it will continue to move at 30 km/h (b/c an object in motion...)

System 2 is the train
Also moving at 30 km/h initially
Let's call the instant the brakes are applied t = 0s
Uniform Acceleration = -2.5 m/s²

As I read it this question is basically asking for the stopping distance of the train. If at the exact instant the car breaks away, the brakes are applied, and the car comes to rest at the back of the train, then the two distances are the same.

Since you have uniform acceleration, use the kinematic equations.

(vf)² - (vi)² = 2ad

30 km/h = 8.33 m/s

d = -(8.33 m/s)²/(2*(-2.5 m/s²)) = 6.94 m

The answer seems low, even for this very slow train, but I can't think of another way to do it. And it seems as if kinematics is the way to go since there's no mention of forces or momentum.
Do you think it's possible to stop a train, even a slow one, in 23 ft? Then again I suppose homework questions aren't that realistic.
Maybe I'm over thinking it...

[dballeng]

Thanks for all input