Answer:
a. 4d.
If the car travels at a velocity of 2v, the minimum stopping distance will be 4d.
Explanation:
Hi there!
The equations of distance and velocity of the car are the following:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position of the car at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
v = velocity of the car at time t.
Let´s find the time it takes the car to stop using the equation of velocity. When the car stops, its velocity is zero. Then:
velocity = v0 + a · t v0 = v
0 = v + a · t
Solving for t:
-v/a = t
Since the acceleration is negative because the car is stopping:
v/a = t
Now replacing t = v/a in the equation of position:
x = x0 + v0 · t + 1/2 · a · t² (let´s consider x0 = 0)
x = v · (v/a) + 1/2 · (-a) (v/a)²
x = v²/a - 1/2 · v²/a
x = 1/2 v²/a
At a velocity of v, the stopping distance is 1/2 v²/a = d
Now, let´s do the same calculations with an initial velocity v0 = 2v:
Using the equation of velocity:
velocity = v0 + a · t
0 = 2v - a · t
-2v/-a = t
t = 2v/a
Replacing in the equation of position:
x1 = x0 + v0 · t + 1/2 · a · t²
x1 = 2v · (2v/a) + 1/2 · (-a) · (2v/a)²
x1 = 4v²/a - 2v²/a
x1 = 2v²/a
x1 = 4(1/2 v²/a)
x1 = 4x
x1 = 4d
If the car travels at a velocity 2v, the minimum stopping distance will be 4d.
