The speed limit in a school zone is 40.0 km/hr. A driver traveling at this speed sees a child run into the road 14 m ahead of the car. The driver applies the brakes and the car slows at a uniform rate of 8.3 m/s^2. If the driver's reaction time is 0.27 sec, what distance does the car travel before stopping?
distance=vi*(t+.27)-1/2 8.3 (t^2)
solve for time t. change vi to m/s
45
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To find the distance the car travels before stopping, we need to consider two parts: the distance covered during the driver's reaction time and the distance covered while braking.
First, let's calculate the distance covered during the driver's reaction time:
1. Convert the reaction time into seconds: 0.27 sec.
2. Multiply the reaction time by the initial speed of the car to get the distance covered during the reaction time: (0.27 s) x (40.0 km/hr) x (1 hr/3600 s) x (1000 m/km) = 3.0 m.
Now, let's calculate the distance covered while braking:
1. The car's initial speed is 40.0 km/hr. Convert this to m/s: (40.0 km/hr) x (1 hr/3600 s) x (1000 m/km) = 11.1 m/s.
2. The car slows down at a uniform rate of 8.3 m/s^2.
3. Use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
4. Rearranging the equation, we get s = (v^2 - u^2) / (2a).
5. Substitute the known values: s = (0 - (11.1 m/s)^2) / (2 * (-8.3 m/s^2)) = 7.7 m.
Finally, add the distance covered during the reaction time and the distance covered while braking to determine the total distance traveled before stopping:
3.0 m + 7.7 m = 10.7 m.
Therefore, the car travels a distance of 10.7 meters before stopping.