Zach is traveling West at a steady 60.0 ml/hr (26.8 m/s) on a straight portion o
ID: 1564085 • Letter: Z
Question
Zach is traveling West at a steady 60.0 ml/hr (26.8 m/s) on a straight portion of Route 1-90. The car directly ahead of him is also travelling at 60 ml/hr. Zach is keeping what he feels is a safe distance from the car directly of him, 27.0 m (roughly 6 car lengths). Suddenly at t = 0, having noticed a posted speed limit of 30.0 ml/hr (due to road construction), the driver of the car ahead on his brakes and begins to show down at a steady rate of 12.0 m/s^2. Once has slowed down to 30.0 ml/hr (13.4 m/s), he plants to maintain this constant speed. Suppose that it takes Zach 1.00 second to notice that the car ahead has slowed down, and another 2.00 seconds to effectively supply his brakes. Also suppose that, once he has applied his brakes, each's car slows down at a steady rate of 12.0 m/s^2. Assume, for the sake of argument, that no collision occurs. What distance would the car ahead travel while slowing down to 30.0 ml/hr? a. 32.4 m. b. 22.4 m. c. 44.9 m. d. 113 m. e, 54.9 m. Which of the following statements is true? a. There is no collision, but at the instant each's car has slowed down to 30.0 ml/hr, his front bumper is approximately 1.0 meters from the rear bumper of the car ahead. b. There is no collision, but at the instant each's car has slowed down to 30.0 ml/hr, his front bumper is approximately 1.0 meter from the rear bumper of the car ahead. c. A collision occurs before the car ahead has slowed down to 30.0 ml/hr. d. A collision occurs after the car ahead has slowed down to 30.0 ml/hr but before zach has even applied his brakes. e. A collision occurs after Zach has applied his brakes but before he can slow down to 30.0 ml/hr.Explanation / Answer
distance x = (vf^2 - vi^2)/(2*a)
acceleration a = -12 m/s^2
distance x = -(13.4^2-26.8^2)/(2*12)
x = 22.4 m
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17)
time taken by car ahead to slow down
t = (vf - vi)/a
t = -(13.4-26.8)/12
t = 1.12 s
distance travelled by Zach before applying brakes
x1 = (26.8*3) = 80.4 m
position of car 49.4 m
x1 > 49.4 m
option c
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