Police find skid marks 66 m long on the highway showing where the car made an emergency stop. Assuming that the acceleration was -10 m/s2 (about the maximum for dry pavement), how fast was the car going?
Vf^2=Vi^2+2ad where Vf=0, solve for Vi
To find out how fast the car was going, we can use the kinematic equation that relates acceleration, initial velocity, final velocity, and distance. The equation is:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity of the car (0 m/s, because it came to a stop)
- vi is the initial velocity of the car (which we need to find)
- a is the acceleration (-10 m/s^2)
- d is the distance (66 m)
Rearranging the equation and plugging in the values, we have:
vi^2 = vf^2 - 2ad
Since vf = 0 and a = -10 m/s^2, the equation simplifies to:
vi^2 = 2 * 10 * 66
vi^2 = 1320
To solve for vi, we need to take the square root of both sides:
vi = √1320
vi ≈ 36.3 m/s
Therefore, the car was going approximately 36.3 m/s before making the emergency stop.