Question #721

A woman driving a car traveling 30 m/sec spots a board in the road. The reaction time of the woman is 0.20 seconds and the car can brake with a deceleration of 4.0 m/sec^2. How far does the car travel before the brakes are applies? What is the total stopping distance from the time she spotted the board? (hint: The total distance is the distance before she applies the brakes afterwards.)

Expert's answer

Before she reacts, it takes 0.2 sec so the car goes x=v*t=30*0.2=6 m while she still hasn't reacted.

Now after she brakes, the cars equation of motion is one with negative constant acceleration and the distance it goes is derived from

v^2 - (v_0)^2 = -2*a*x => x= [v^2 - (v_0)^2]/(-2a)

Since v = final velocity = 0 we get

x= [- (v_0)^2]/(-2a) =(v_0)^2/(2a) = 30^2/(2*4)=900/4= 225 m

Now we should add this to the distance the car goes before she reacts so the total distance is

x=225+6=231 m

Now after she brakes, the cars equation of motion is one with negative constant acceleration and the distance it goes is derived from

v^2 - (v_0)^2 = -2*a*x => x= [v^2 - (v_0)^2]/(-2a)

Since v = final velocity = 0 we get

x= [- (v_0)^2]/(-2a) =(v_0)^2/(2a) = 30^2/(2*4)=900/4= 225 m

Now we should add this to the distance the car goes before she reacts so the total distance is

x=225+6=231 m

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