You may skip these questions entirely, answer them in addition to questions in t

Question

You may skip these questions entirely, answer them in addition to questions in the main body of the exam, or answer them in place of questions in the main body of the exam. I will simply add up the number of correct answers on your scantron form, divide by 20, and multiply by 100% to obtain your exam score. It is possible to earn 115% on this exam! Maggie is driving at a constant 70.0 mi/hr (31.3 m/s) along a straight section of Route 3 North. The car directly ahead of her is also moving at 7o.o mi/hr. The front bumper of Maggie's car is located 25.0 m from the rear bumper of the car ahead when, at t = 0, the driver of the car ahead applies his brakes and begins to slow down at a steady rate of 6.00m.s^2, towards the posted "construction zone" speed limit of 40.0 mi/hr (17.9 m/s). After slowing to 40.0 mi/hr, the driver of the car ahead plans to maintain this constant speed. It takes Maggie 2.00 seconds to notice that the car ahead is slowing down, and another 1.00 second to effectively apply her brakes and begin to slow down. Assume that, once Maggie effectively applies the brakes, her car sows down at a steady rate of 6.00 m/s^2. At t = 0, the Maggie's tort bumper is located at X = 0 and the car ahead begins to slow down at a steady 600 m/s^2. Assume, for the sake of argument, that no collision occurs, Starting at t = 0, what distance does car ahead travel while slowing down to 40.0 mi/hr? a. 51.7m. b. 550 m. c. 54.9 m. d. 275 m. e. 110m. Assume, for the sake of argument, that no collision occurs. At the instant when the car ahead has slowed to 40 mi/hr, the front bumper of Maggie's car is located at x equals a 29.9 m. b. 40.0 m. c. 89.3 m. d. 69.9 m. e. 156 m. Assume, for the sake of argument, that no collision occurs. At the instant when Maggie effectively applies her brakes, the rear bumper of the car ahead would be located at x equals a. 93.7 m. b. 119 m. c, 331 m. d. 68.7m. e. 79.9 m.

Explanation / Answer

21. vi = 70 mi/h = 70 x 1609 m / 3600s = 31.3 m/s

vf = 40 mph = 40 x 1609 / 3600 = 17.9 m/s

a = - 6 m/s^2

Applying vf^2 - vi^2 =2 a d

17.9^2 - 31.3^2 = 2(-6) (d)

d = 54.9 m
Ans(c)

22. time taken to slow down ,

vf = vi + at

17.9 = 31.3 - 6t

t = 2.23 s

for 3 sec. constant speed of 31.3 m/s

d1 = 31.3 x 2.23 = 69.8 m

Ans(d)

23. maggies applies brake after 3 sec,

distance traveled in 2.23 sec

d1 = 54.9 m

after that for 3 - 2.23 = 0.77 s

d2 = 17.9 x 0.77 = 13.8 m

x = d1 + d2 + 25

x = 93.7 m

Ans(a)

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