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?
d = (V^2-Vo^2)/2a = (25-0) / 2.5 = 10 m.
6.25
To determine the distance the boy would reach a speed of 5.00 m/s while sledding down the hill, we can use the kinematic equation:
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
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Substituting the given values into the equation:
s = (5.00^2 - 0^2) / (2 * 1.25)
s = (25.00 - 0) / 2.50
s = 10.00
Therefore, the boy would reach a speed of 5.00 m/s after traveling a distance of 10.00 meters down the hill.
To find the distance the boy would cover, we can use the equation of motion:
\(v^2 = u^2 + 2as\)
Where:
- \(v\) is the final velocity (5.00 m/s)
- \(u\) is the initial velocity (0 m/s, as the boy started from rest)
- \(a\) is the acceleration (1.25 m/s²)
- \(s\) is the distance traveled
We rearrange the equation to solve for distance (\(s\)):
\(s = (v^2 - u^2) / (2a)\)
Plugging in the values we have:
\(s = (5.00^2 - 0^2) / (2 * 1.25)\)
Simplifying this equation:
\(s = 25.00 / 2.50\)
\(s = 10.00\) meters
Therefore, the boy would reach a speed of 5.00 m/s in a distance of 10.00 meters down the hill.