A boy sledding down a hill accelerates at 1.25 m/s2. If he started from rest, in what distance would he reach a speed of 5.00 m/s?
v = a t
5 = 1.25 * t so t = 4
d = (1/2) a t^2 so d = (1/2)(1.25)(16)
To find the distance the boy would reach a speed of 5.00 m/s, we can use the kinematic equation that relates acceleration, initial velocity, final velocity, and distance:
v^2 = u^2 + 2as
Where:
v = final velocity (5.00 m/s)
u = initial velocity (0 m/s, since the boy starts from rest)
a = acceleration (1.25 m/s^2)
s = distance (what we want to find)
We can rearrange the equation to solve for s:
s = (v^2 - u^2) / (2a)
Substituting the given values:
s = (5.00^2 - 0^2) / (2 * 1.25)
Simplifying the equation:
s = (25.00) / (2.50)
s = 10.00 meters
Therefore, the boy would reach a speed of 5.00 m/s after traveling a distance of 10.00 meters.