Achild accidentally runs into the street in front of an approaching vehicle. the vehicle is traveling at 40 mi/h. assuming the road is level, at what distance must the driver first see the child to stop just in time?

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.

higher potential

explanation:

because it is in a high speed so it must move into a of high potential