The formula below can be used to find the distance (d) traveled by an object, where v represents the initial velocity of the object, t represents the time, in seconds, and d represents the acceleration, in meters per second squared.
d=vt+1/2at with a a square route of 2
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 squared for 12 seconds. What is the distance, in meters, that the bicycle travels?
v=4.5m/s
a=0.4m/s
t=12s
d= vt+1/2at^2
d= 4.5(12) + 1/2(0.4)(12)^2
d= 54+28.8
d= 82.8m
To find the distance traveled by the bicycle, we can use the formula:
d = vt + (1/2)at^2
Given values:
v = 4.5 m/s (initial velocity)
a = 0.4 m/s^2 (acceleration)
t = 12 s (time)
First, substitute the given values into the formula:
d = (4.5 m/s)(12 s) + (1/2)(0.4 m/s^2)(12 s)^2
Now, let's simplify the equation step-by-step:
d = 54 m + (1/2)(0.4 m/s^2)(144 s^2)
d = 54 m + (1/2)(0.4 m/s^2)(144)
d = 54 m + (1/2)(0.4)(144)m
Next, calculate the result within parentheses:
d = 54 m + (0.2)(144)m
d = 54 m + 28.8 m
Finally, add the two values together to find the total distance traveled:
d = 82.8 m
Therefore, the bicycle travels a distance of 82.8 meters down the hill.