A driver has a reaction time of 0.50 s , and the maximum deceleration of her car is 6.0 m/ s 2 . She is driving at 20 m/s when suddenly she sees an obstacle in the road 50 m in front of her. what is the distance she passes after noticing the obstacle before fully stopping?
The correct answer and explanation is:
Correct Answer: 53.3 meters
Explanation:
To find the total distance the driver travels after noticing the obstacle before coming to a complete stop, we divide the motion into two parts:
Part 1: Distance traveled during reaction time
During the reaction time of 0.50 seconds, the driver continues to move at the same initial speed of 20 m/s, since no braking has occurred yet.
Using the formula for constant velocity: d1=v⋅td_1 = v \cdot t d1=20 m/s⋅0.50 s=10 metersd_1 = 20 \, \text{m/s} \cdot 0.50 \, \text{s} = 10 \, \text{meters}
Part 2: Distance traveled while decelerating
Once the driver reacts, she begins to decelerate at 6.0 m/s² until the car stops. We calculate the braking distance using the kinematic equation: v2=u2+2adv^2 = u^2 + 2ad
Here:
- Final velocity v=0 m/sv = 0 \, \text{m/s}
- Initial velocity u=20 m/su = 20 \, \text{m/s}
- Acceleration a=−6.0 m/s2a = -6.0 \, \text{m/s}^2 (deceleration)
Solving for dd: 0=(20)2+2(−6.0)d0 = (20)^2 + 2(-6.0)d 0=400−12d0 = 400 – 12d 12d=400⇒d=40012≈33.3 meters12d = 400 \Rightarrow d = \frac{400}{12} \approx 33.3 \, \text{meters}
Total stopping distance: Total distance=d1+d2=10+33.3=43.3 meters\text{Total distance} = d_1 + d_2 = 10 + 33.3 = 43.3 \, \text{meters}
Comparison with obstacle distance:
The obstacle is 50 meters away, and the driver stops after 43.3 meters. This means she does not hit the obstacle and stops safely with about 6.7 meters to spare.
This problem illustrates how reaction time and deceleration rate are critical in stopping distances. Even a small delay in reaction can significantly increase the total distance traveled before stopping.