A typical car under hard braking loses speed at a rate of about 6 m/s2. The typical reaction time to engage the brakes is 0.50 s. A local school board sets the speed limit in a school zone such that all cars should be able to stop in 5.0 m. What maximum speed does this imply for a car in the school zone?
goes .5 Vi in the first half second
for the rest of the trip the average speed is Vi/2 (because Vi at start, zero at end, and constant a)
5 = .5 Vi + (Vi/2) t =
t is the time to deaccelerate from Vi to 0
v = Vi - 6 t
so t = Vi/6
so
5 = .5 Vi +(Vi/2)(Vi/6)
Vi^2/12 + Vi/2 - 5 = 0
Vi^2 + 6 Vi - 60 = 0
Vi = [ -6 +/- sqrt(36+240)]/2
Vi = 10.6/2
Vi = 5.3 m/s
=================check
5.3 m/s * .5 = 2.6 meters before braking
t = Vi/6 = .88 second
d = 5.3*.88/2 = 2.3 m
2.6+2.3 = 4.9, close enough
To determine the maximum speed for a car in the school zone, we need to calculate the stopping distance and then use it to find the maximum speed.
First, let's calculate the stopping distance using the given information:
Initial speed (u) = unknown
Final speed (v) = 0 (as the car comes to a stop)
Deceleration (a) = -6 m/s^2 (negative sign indicates deceleration)
Reaction time (t) = 0.50 s
The formula to calculate stopping distance is:
s = (u * t) + (0.5 * a * t^2)
Plugging in the values, we get:
5.0 m = (u * 0.50 s) + (0.5 * -6 m/s^2 * (0.50 s)^2)
Simplifying the equation:
5.0 m = (0.50 u) - 0.75 m
0.75 m = 0.50 u
u = (0.75 m) / (0.50 s)
u = 1.5 m/s
Now that we know the initial speed (u) required to stop within the given distance, we can calculate the maximum speed:
Maximum speed = initial speed (u)
Maximum speed = 1.5 m/s
Therefore, the maximum speed for a car in the school zone should be 1.5 m/s.
To determine the maximum speed allowed in the school zone, we need to consider the distance it takes for a car to stop and the reaction time to engage the brakes.
Let's denote the maximum speed as v in m/s and the total stopping distance as s in meters.
According to the problem, the total stopping distance is 5.0 m, which can be divided into two parts: the distance traveled during the reaction time and the distance traveled while decelerating.
The formula for the distance traveled during the reaction time is given by:
Distance_react = (initial velocity) * (reaction time)
The distance traveled while decelerating can be calculated using the equation of motion:
Distance_decel = (initial velocity * deceleration) / 2
Since the car is under hard braking, the deceleration is given as 6 m/s^2.
The total stopping distance is the sum of the distance traveled during the reaction time and the distance traveled while decelerating:
s = Distance_react + Distance_decel
Substituting the formulas for Distance_react and Distance_decel, we get:
5.0 = (v * 0.50) + (v^2 / (2 * 6))
Multiplying through by 6 to eliminate the fraction, we have:
30 = 3v + (v^2 / 12)
Rearranging the equation:
v^2 / 12 + 3v - 30 = 0
Now we can solve this quadratic equation to find the maximum speed:
Using the quadratic formula, which states that for an equation in the form ax^2 + bx + c = 0, the solutions for x can be calculated using:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1/12, b = 3, and c = -30. Substituting these values into the quadratic formula:
v = (-3 ± √((3^2) - 4 * (1/12) * -30)) / (2 * (1/12))
Simplifying:
v = (-3 ± √(9 + 10)) / (2/12)
v = (-3 ± √19) / (1/6)
Multiplying the top and bottom by 6 to get rid of the fraction:
v = (-3 ± √19) * 6
Now we can calculate the maximum speed by evaluating both the positive and negative solutions:
v₁ = (-3 + √19) * 6
v₂ = (-3 - √19) * 6
Therefore, the maximum speed allowed in the school zone is approximately v₁ and v₂ m/s, respectively.