in that first half second she goes 10 meters, 40 meters left
then
v = 20 - 6 t
at stop v = 0 so
t = 20/6
average speed during stop = (20 + 0) /2 = 10 m/s
so goes 10 *20/6 more = 100/3 = 33 1/3 meters
well, we had 40 :)
A driver has a reaction time of 0.50 s, and the maximum deceleration of her car is 6.0 m/s squared. She is driving at 20 m/s when suddenly she sees an obstacle in the road 50 m in front of her. Can she stop the car to avoid the collision?
1 answer