Answer:
The driver must see the child at a distance of 22.8 m to avoid hitting him.
Explanation:
Hi there!
According to this paper, "Nerijus Kudarauskas (2007) Analysis of emergency braking of a vehicle, Transport, 22:3, 154-159", the braking acceleration of a car with ABS is about 9 m/s².
My reaction time is about 0.28 s (tested online on https://faculty.washington.edu/chudler/java/redgreen.html)
The traveled distance during the time I take to react can be calculated as follows:
x = v · t
Where:
x = traveled distance.
v = velocity.
t = time.
40 mi/h = 17.9 m/s
x = 17.9 m/s · 0.28 s
x = 5.01 m
Now, let´s calculate the time it takes the car to stop using the equation of velocity:
v = v0 + a · t
Where:
v = velocity.
v0 = initial velocity.
a = acceleration.
t = time.
When the car stops, its velocity is zero. The initial velocity is 17.9 m/s and the maximum brake decceleration is 9 m/s²:
0 m/s = 17.9 m/s - 9 m/s² · t
-17.9 m/s / - 9 m/s² = t
t = 2.0 s
The distance traveled while braking will be:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = traveled distance.
x0 = initial position.
v0 = initial velocity.
t = time.
a = accleration.
Considering the initial position as the position at which the brakes are applied, x0 = 0.
x = 0 + 17.9 m/s · 2.0 s - 1/2 · 9 m/s² · (2.0 s)²
x = 17.8 m
Then, the car will travel (17.8 m + 5.01 m) 22.8 m before stopping.
The driver must see the child at a distance of 22.8 m to avoid hitting him.
