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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).

User Jdabrowski
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Answer:


1.875\ \text{m}


1.5625\ \text{m}

Step-by-step 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=(v^2-u^2)/(2a)\\\Rightarrow s=(0-30)/(2* -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=(0-25)/(2* -8)\\\Rightarrow s=1.5625\ \text{m}

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

User Dlanced
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