A car with an initial speed of 80 km/h is brought to a stop by applying a negative acceleration of 0.2g.  What is the stopping distance of the car? If a Car with good brakes, but with bad tires maximaly reach a acceleration or retardation of 0,2g, without slipping (g=gravitational acceleration). The car itself drives at 80km/h(or 49.7096954 mph). how big is the stopping distance then ?  

Expert Answers

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To solve this problem we must first have a unified set of units.

If g = 9.80 m/s^2  then we must change the units of speed to m/s:

80 km/h[(1m/s)/(3.6km/h)] = 22.2 m/s

We can then apply the equation

V^2 = Vi^2 + 2AD

The car is coming to a stop so V = 0

A = -(.2)g = -1.96 m/s^2

Vi^2 = -2AD

D = -Vi^2/2A

D = -(22.2m/s)^2/(-3.92m/s^2)

D = 126 m

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