A boy sledding down a hill accelerates at 1.20 m/s2. If he started from rest, in what distance would he reach a speed of 7.00 m/s?
V^2 = Vo^2 + 2a*d
d = (V^2-Vo^2)/2a
d = (7^2-0)/2.4
Use:
v(t) = v(0) + a t
With values:
(7.00m/s) = (0.00m/s) + (1.20m/s^2) t
Solve for t.
To find the distance the boy would reach a speed of 7.00 m/s, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity = 7.00 m/s
u = initial velocity = 0 m/s (because the boy started from rest)
a = acceleration = 1.20 m/s^2
s = distance
Rearranging the equation gives:
s = (v^2 - u^2) / (2a)
Substituting the given values:
s = (7.00^2 - 0^2) / (2 * 1.20)
s = (49.00 - 0) / 2.40
s = 20.42 m
Therefore, the boy would reach a speed of 7.00 m/s after traveling a distance of approximately 20.42 meters.
To find the distance the boy would reach a speed of 7.00 m/s, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
- v is the final velocity (7.00 m/s)
- u is the initial velocity (0 m/s, as the boy started from rest)
- a is the acceleration (1.20 m/s^2)
- s is the distance traveled (what we are trying to find)
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Substituting the given values:
s = (7.00^2 - 0^2) / (2 * 1.20)
Calculating this expression, we find:
s = 24.92 meters
Therefore, the boy would reach a speed of 7.00 m/s after traveling a distance of 24.92 meters.