the relationship b/w the distance d, in feet, required to stop a vehicle and s, the speed in miles per hour that the vehicle was traveling, is given by the equation
d=0.0155s(squared)/f
f represents the coefficient of friction b/w the tires and the road
*it took a car 205 feet to stop. what speed was the car traveling? use f=0.3 and round your answer to the nearest mile per hour.
i did 205= 0.0155s(squared)/0.3...idk how to solve this
205 = 0.0155 s^2 / 0.3
multiply both sides by 0.3
61.5 = 0.0155 s^2
divide both sides by 0.0155
3968 = s^2
take the square root of both sides
63 miles per hour
To solve the given equation, follow these steps:
1. Start with the given equation: d = 0.0155s^2 / f
Replace d with 205 (the distance the car took to stop) and f with 0.3 (the coefficient of friction):
205 = 0.0155s^2 / 0.3
2. Multiply both sides of the equation by 0.3 to get rid of the denominator:
0.3 * 205 = 0.0155s^2
61.5 = 0.0155s^2
3. Divide both sides of the equation by 0.0155 to isolate s^2:
61.5 / 0.0155 = s^2
3967.74 = s^2
4. Take the square root of both sides to solve for s:
√3967.74 = √s^2
s = ± 62.99
As the speed of the car cannot be negative, we take the positive value:
s = 62.99 mph
5. Round the answer to the nearest whole number as instructed:
Rounded to the nearest mph, the speed of the car was 63 mph.
Therefore, the speed of the car was approximately 63 miles per hour.