Final answer:
To stop a 1000 kg car traveling at 25 m/s, with a coefficient of static friction of 0.18, it will take approximately 14.17 seconds. This is calculated using the relationship between force, mass, acceleration, and time.
Step-by-step explanation:
To calculate how long it will take for a 1000 kg car traveling at 25 m/s to stop when the brakes are applied, we must first determine the force of friction acting on the car, which is what ultimately brings the car to a stop. The coefficient of static friction is given as 0.18. The force of friction (F_friction) can be calculated using the formula F_friction = μ * N, where μ is the coefficient of friction and N is the normal force. For a car on a flat surface, the normal force (N) is equal to the weight of the car, which is the mass (m) times the gravitational acceleration (g), so N = m * g.
Assuming g is 9.8 m/s², we can calculate the frictional force as follows:
F_friction = 0.18 * 1000 kg * 9.8 m/s² = 1764 N
The deceleration (a) of the car can be calculated using Newton's second law, which states that F = m * a. Rearranging the formula for acceleration gives us a = F/m. Plugging in the values, we get:
a = 1764 N / 1000 kg = 1.764 m/s²
To find the time (t) it takes for the car to stop, we use the equation v = a * t, where v is the initial velocity. Rearranging for t, we have t = v/a. Substituting the given values:
t = 25 m/s / 1.764 m/s² = 14.17 s
Therefore, it will take approximately 14.17 seconds for the car to come to a complete stop.