Time, Distance & Speeds

Fundamental Relationships:

1. An easy one: Distance = Speed x Time -Example: 30 mph x 10 hrs = 300 miles

2. A useful one: 1 mph = 5280 ft/hr /3600 sec/hr = 1.47 ft/sec. So: 20 mph = 29.4 ft/sec.

3. Acceleration: The distance covered while accelerating for T secs.:

D = Vo x T + 1/2 A x T xT where: Vo is the vehicle's speed when it starts accelerating ( 0 for a standing start) and A is the acceleration rate- for cars this is about 10 ft/sec/sec.

Example: Vo = 0, T = 2 sec. D = 1/2 x 10 x 2 x 2 = 20 ft.

The time required to cover a given distance from a standing start:

T = sqrt ( 2 x D / A ) - Many calculators will do square roots.

4. Braking Distance: The total distance required to stop the vehicle equals the distance covered while the driver is detecting and reacting to the situation plus the distance covered while braking. So D = Vo x T + the braking distance :

Db = S x S /( 30 x Cd ) where:

S is the speed in mph and Cd is the drag or braking coefficient, 0.7 - 0.8.

Braking time: Take the speed in ft/sec and devide it by the deceleration rate- about 0.7 -0.8 x 32.2 = 22- 26 ft/sec/sec. Add the reaction time.

5. Speed from skid marks:

If the vehicle is stopped at the end of the skid marks, then the original speed is : S = 5.5 x sqrt ( Cd x Length of skid marks) where the Cd is as defined above and the speed is in mph. If the vehicle is still moving when the skid marks end, this formula does not yield the original speed. (see the "Combined Speed" sheet)

6. Speed from damage:

It is possible to compute speed changes from crush damage but the calculations are complicated and generally require a computer along with damage measurements and knowledge of the structural properties of the vehicles involved.

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