# 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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