Question

For a given motor vehicle, the maximum achievable deceleration from braking is approximately 8 m/sec2 on dry concrete. On wet asphalt, it is approximately 2.5 m/sec2. Given that 1 mph corresponds to 0.447 m/sec, find the total distance that a car travels in meters on dry concrete after the brakes are applied until it comes to a complete stop if the initial velocity is 67 mph (30 m/sec) or if the initial braking velocity is 56 mph (25 m/sec).

Answer 1

Answer:

$1.875\ \text{m}$

$1.5625\ \text{m}$

Explanation:

$a$ = Acceleration on dry concrete = $-8\ \text{m/s}^2$

$v$ = Final velocity = 0

$u$ = Initial velocity

$s$ = Displacement

For $u=30\ \text{m/s}$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-30}{2\times -8}\\\Rightarrow s=1.875\ \text{m}$

The distance traveled to come to a stop is $1.875\ \text{m}$

For $u=25\ \text{m/s}$

$s=\dfrac{0-25}{2\times -8}\\\Rightarrow s=1.5625\ \text{m}$

The distance traveled to come to a stop is $1.5625\ \text{m}$.