A car is traveling at 60.6 mph on a horizontal highway. The acceleration of gravity is 9.8 m/s2. If the coefficient of friction between road and tires on a rainy day is 0.056, what is the minimum distance in which the car will stop? Answer in units of m.
Vf^2=vi^2+2ad
but a= -mu*mg/m= -mu*g
solve for d
Why did the car bring an umbrella to the race? Because it wanted to be prepared for the rainy day! Now, let's calculate the minimum distance the car will stop.
We'll convert the car's speed from miles per hour to meters per second. 60.6 mph is approximately 27.0556 m/s. Now, let's use this information to find the deceleration of the car.
The frictional force opposing the car's motion is given by the equation:
Frictional force = (coefficient of friction) * (normal force)
The normal force is the weight of the car, which is given by:
Normal force = mass * gravitational acceleration
The frictional force is also equal to mass * acceleration, so we have:
mass * acceleration = (coefficient of friction) * (mass * gravitational acceleration)
Now, we can cancel out the mass:
acceleration = (coefficient of friction) * (gravitational acceleration)
Plugging in the values, we have:
acceleration = 0.056 * 9.8 m/s^2
Now, let's find the time it takes for the car to stop. We use the equation:
Final velocity = Initial velocity + (acceleration) * (time)
Since the final velocity is 0 m/s (since the car stops), we can solve for the time:
0 = 27.0556 m/s + (-0.056 * 9.8 m/s^2) * (time)
Now, we can solve for time:
time = -27.0556 m/s / (-0.056 * 9.8 m/s^2)
Finally, we can find the minimum distance the car will stop by using the equation:
Distance = (Initial velocity) * (time) + (1/2) * (acceleration) * (time^2)
Substituting the values, we can calculate the minimum distance.
To find the minimum distance in which the car will stop, we will use the equation of motion that relates distance, initial velocity, acceleration, and coefficient of friction.
The equation is:
v^2 = u^2 + 2as
where:
v = final velocity (which is 0 in this case, as the car is stopping)
u = initial velocity
a = acceleration (coefficient of friction * acceleration due to gravity)
s = distance
Converting the initial velocity from mph to m/s:
u = 60.6 mph * (1.60934 km/h) * (1 h / 3600 s) = 27.08 m/s
Substituting the values into the equation:
0^2 = (27.08)^2 + 2 * (0.056) * (9.8) * s
Simplifying,
0 = 732.6464 + 10.8352s
Rearranging the equation to solve for s,
10.8352s = -732.6464
s = -732.6464 / 10.8352
s ≈ -67.58
Since distance cannot be negative, the minimum distance in which the car will stop is approximately 67.58 meters.
To find the minimum distance in which the car will stop, we need to calculate the stopping distance using the equation:
Stopping distance = (initial velocity^2) / (2 * acceleration * coefficient of friction)
Now, let's plug in the given values into the equation:
Initial velocity (v) = 60.6 mph = 60.6 * 0.44704 m/s (since 1 mph = 0.44704 m/s)
= 27.107424 m/s
Acceleration (a) = acceleration of gravity = 9.8 m/s^2
Coefficient of friction (μ) = 0.056
Now, substitute these values into the formula:
Stopping distance = (27.107424^2) / (2 * 9.8 * 0.056)
= 734.59846336 / 1.09
≈ 674.534 meters (rounded to three decimal places)
Therefore, the minimum distance in which the car will stop is approximately 674.534 meters.