Question

If a car’s speed is 70 miles per hour on a road surface with a drag factor of 0.50 and with a breaking efficiency of 85% (n = .85), what would be the length of the skid marks? Round to the nearest tenth.

Answer 1

Answer:

About 384.3 feet

Step-by-step explanation:

The formula for that is given by,

$S= \sqrt{30 \times d \times f \times n}$

Where,

S is the speed of the car

d is the distance of skid

f is the drag factor for the road surface

n is breaking efficiency as percentage

Putting the values we get,

$S^2= 30 \times d \times f \times n$

$d= \frac{S^2}{30 \times f \times n}$

$d = \frac{70^2}{30 \times 0.50 \times 0.85} = 384.3$ feet

So the length of skid marks would be 384.3 feet.