biker passes a lamppost at the crest of a hill at + 4.5 m/s. she acceleratets down the hill at a constant rate of +0.40 ,/s2 for 12 seconds.\ how far does she move down the hill during this time?..

83 m/s

To find the distance the biker moves down the hill, we can use the equation of motion:

d = v_i * t + (1/2) * a * t^2

Where:
- d is the distance,
- v_i is the initial velocity,
- a is the acceleration, and
- t is the time.

In this case:
- The initial velocity (v_i) is +4.5 m/s,
- The acceleration (a) is +0.40 m/s^2, and
- The time (t) is 12 seconds.

Now, substituting these values into the equation, we have:

d = (4.5 m/s) * (12 s) + (1/2) * (0.40 m/s^2) * (12 s)^2

Calculating this equation:
d = 54 m + 0.40 m/s^2 * 144 s^2
d = 54 m + 57.6 m
d = 111.6 m

Therefore, during this time, the biker moves down the hill by approximately 111.6 meters.