The formula below can be used to find the distances (d) traveled by an object, where v represents the initial velocity of the object, t represents the time, in seconds, and a represents the acceleration, in meters per second squared.

d=vt+1\2at2

A bicycle traveling at a velocity of 4.5 meters per second accelerates down a hill at a constant rate of 0.4 meters per second sqaured for 12 seconds. What is the distance, in meters, that the bicycle travels?

To find the distance traveled by the bicycle, we can use the formula:

d = vt + (1/2)at^2

Given values:
v = 4.5 meters per second (initial velocity)
a = 0.4 meters per second squared (acceleration)
t = 12 seconds (time)

Let's substitute the values into the formula and calculate the distance:

d = (4.5)(12) + (1/2)(0.4)(12)^2
= 54 + (1/2)(0.4)(144)
= 54 + 28.8
= 82.8 meters

Therefore, the bicycle travels a distance of 82.8 meters.

To find the distance traveled by the bicycle, we can use the formula:

d = vt + (1/2)at^2

Given:
v = 4.5 meters per second (initial velocity of the bicycle)
a = 0.4 meters per second squared (acceleration of the bicycle down the hill)
t = 12 seconds (time for which the bicycle accelerates)

Substituting the values into the formula:

d = (4.5)(12) + (1/2)(0.4)(12^2)

First, calculate (1/2)(0.4)(12^2):

(1/2)(0.4)(12^2) = 0.2 * 0.4 * 144 = 0.2 * 57.6 = 11.52

Next, calculate (4.5)(12):

(4.5)(12) = 54

Finally, add the two results together:
d = 54 + 11.52 = 65.52 meters

Therefore, the bicycle travels approximately 65.52 meters down the hill.