Answer:
Please see the explanation.
Explanation:
Hi there!
During the reaction time, the car travels at a constant velocity and its position is calculated as follows:
x = v · t
Where:
x = position of the car.
v = velocity.
t = time
In our case:
v = v0
t = tR
Then, the distance traveled by the car before aplying the brakes will be:
x = v0 · tR
While the car is stopping, the equation of the position is following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x0 = initial position.
a = acceleration.
t = time.
v0 = intial velocity
In our case, the initial position, x0, is the distance traveled during the reaction time: x0 = v0 · tR
So, the equation of position will be:
X = v0 · tR + v0 · t + 1/2 · a · t²
Let´s find the time it takes the car to stop while slowing down with a deceleration "a". For this, we have to use the equation of velocity (v):
v = v0 + a · t
When the car stops, v = 0. Then:
0 = v0 + a · t
Solving for "t":
-v0 / a = t
Now, let´s replace t in the equation of position:
X = v0 · tR + v0 · t + 1/2 · a · t²
X = v0 · tR + v0 · (-v0 / a) + 1/2 · a · (-v0/a)²
X = v0 · tR - v0²/a + 1/2 · a · v0²/a²
X = v0 · tR - v0²/a + 1/2 · v0²/a
X = v0 · tR - 1/2 · v0²/a
or written in a different way:
X = v0 · tR - v0²/(2a)