Answer:
The minimum distance, <em>d</em>, between the cars so as to avoid collision is approximately 5.1304 feet
Explanation:
The given parameters are;
The distance of car B ahead of car A = d
The initial speed of both car A and car B = 80 ft./s
The deceleration of car B due to the application of the breaks = 10 ft./s²
The reaction time of the driver of car A = 0.65 s
The deceleration of the driver of car A = 17 ft./s²
To avoid collision, we have;
v² = u² - 2·a·s
Where;
v = The final velocity
u = The initial velocity
a = The acceleration
s = The distance covered
Where the car <em>B</em> comes to rest, we have, v = 0, therefore;
0² = u² - 2·a·s
u² = 2·a·s
s = u²/(2·a)
The distance travelled by car <em>B</em> before coming to rest, <em>s</em>, is therefore;
s = (80 ft./s)²/(2 × 10 ft./s²) = 320 ft.
The distance travelled by car A, after car B applies brakes, is given as follows;
The distance travelled at constant speed = u × Reaction time
∴ The distance travelled at constant speed = 80 ft./s × 0.65 s = 52 ft.
The distance, <em>s</em>, travelled by car A under deceleration is given as follows;
s = (80 ft./s)²/(2 × 17 ft./s²) = 188.24 ft.
s = 80·(t - 0.65) - (1/2)·17·(t - 0.65)² + 80 × 0.65 = d + 80·t - (1/2)·10·t²
∴ d = 80·(t - 0.65) - (1/2)·17·(t - 0.65)² + 80 × 0.65 - (80·t - (1/2)·10·t²)
d = -3.5·t² + 11.05·t - 3.59125
dd/dt = -7·t + 11.05 = 0
The time for the maximum distance for the cars to collide, t = 11.05/7
The maximum distance between the cars for collision, d = -3.5 × (11.05/7)² + 11.05 × (11.05/7) - 3.59125 ≈ 5.1304
Therefore, for no collision, the distance between the cars, <em>d</em>, should be approximately more than 5.1304 feet
