a bike passes a lamp post at the crest of hill at 4.5 m/s. she accelerates down the hill at the rate of 0.04 m/s squared for 12 seconds. how far does she move down the hill during this time ?
s=v₀t+at²/2,
v₀= 4.5 m/s,
a= 0.04 m/s²,
t= 12s
d=V(initial)*time+1/2at^2
d=(4.5)(12)=1/2(0.40)(12)
d=54+1/2(4.8)
d=54+2.4
d=56.4m
To find out how far the bike moves down the hill during this time, we can use the formula for distance traveled with constant acceleration:
distance = initial velocity * time + (1/2) * acceleration * time^2
Given:
Initial velocity (u) = 4.5 m/s
Time (t) = 12 s
Acceleration (a) = 0.04 m/s^2
Plugging in the values into the formula, we get:
distance = (4.5 m/s * 12 s) + (1/2) * (0.04 m/s^2) * (12 s)^2
First, let's calculate the initial velocity term:
(4.5 m/s * 12 s) = 54 m
Now, let's calculate the second term:
(1/2) * (0.04 m/s^2) * (12 s)^2 = 0.5 * 0.04 m/s^2 * 144 s^2
Simplifying:
(0.5 * 0.04 * 144) m^2/s^2 = 2.88 m
Now, let's add the two terms together:
distance = 54 m + 2.88 m
Therefore, the bike moves down the hill for a distance of 56.88 m during this time.