- #1
Heidi Henkel
- 65
- 1
A car going 25 mph with studded tires on packed powder stops in 20 meters. What will be the stopping distance if it is going 20 mph?
We do not know the mass of the car, but perhaps it is accurate to create an acceleration constant that will apply for other speeds?
a=v/t and v=d/t so therefore by combining these we get a=v^2/d
25mph is 11.176 m/s
deceleration constant: a = (11.176m/s)^2 / 20m = 6.245 m/s^2
OK so this is the deceleration constant for that set of tires on that car on that surface...
then can we apply the deceleration constant to the other question? What will be the stopping distance at 20 mph?
20mph = 8.94 m/s; stopping distance is 12.8m
Is this valid?
We do not know the mass of the car, but perhaps it is accurate to create an acceleration constant that will apply for other speeds?
a=v/t and v=d/t so therefore by combining these we get a=v^2/d
25mph is 11.176 m/s
deceleration constant: a = (11.176m/s)^2 / 20m = 6.245 m/s^2
OK so this is the deceleration constant for that set of tires on that car on that surface...
then can we apply the deceleration constant to the other question? What will be the stopping distance at 20 mph?
20mph = 8.94 m/s; stopping distance is 12.8m
Is this valid?