1. Allen is driving North on Highway 69 at 90 km/h and sees a large moose on the road. He quickly slams on his brakes, but his reaction time is 0.85 s (as he sees the moose, thinks
about his response, and then presses the brake pedal). He presses the brake for 3.5 s and
comes to a stop just in time.
a) Find the distance travelled after seeing the moose and before pressing the brake.
b) Find the total distance he travelled before coming to a stop.
c) Find the average acceleration once he presses the brake.

Take the moment Allen sees the moose to be the origin.

First, convert his speed to m/s.

90 km/h = (90 km/h) • (1000 m/km) • (1/3600 h/s) = 25 m/s

(a) For the time it takes him to react (0.85 s), Allen is moving at a constant speed of 25 m/s, so that before he actually does anything, he covers a distance of

(25 m/s) • (0.85 s) = 21.25 m

(b) Once he presses the brakes, Allen's vehicle covers a distance x in time t of

x = 21.25 m + (25 m/s) t + 1/2 a t²

and has a speed v of

v = 25 m/s + a t

It takes him 3.5 s to come to a full stop. Use this to find the acceleration:

0 = 25 m/s + a (3.5 s)

a = - (25 m/s) / (3.5 s)

a ≈ - 7.1 m/s²

After 3.5 s, he will have traveled a total distance of

x = 21.25 m + (25 m/s) (3.5 s) + 1/2 (- 7.1 m/s²) (3.5 s)²

x = 152.5 m ≈ 150 m

(c) This one is worded a bit strangely, specifically "once he presses the brake" seems to suggest instantaneous acceleration, not average. Average acceleration is defined for some duration of time. You're probably expected to report the acceleration of the car as it comes to a stop, which we found earlier to be

a ≈ - 7.1 m/s²

your answer is, orbit around the nucleus