Q1 Car B is traveling a distance d ahead of car A. Both cars are traveling at 80 ft/s when the driver of B suddenly

applies the brakes, causing his car to decelerate at 10 ft/s?. It

takes the driver of car A 0.65 s to react (this is the normal

reaction time for drivers). When he applies his brakes, he

decelerates at 17 ft/s?. Determine the minimum distance d

be tween the cars so as to avoid a collision.

d

The minimum distance, d, 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 B 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 B before coming to rest, s, 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, s, 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, d, should be approximately more than 5.1304 feet

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explanation: