A boy sledding down a hill accelerates at 1.30 m/s2. If he started from rest, in what distance would he reach a speed of 8.00 m/s?
vf^2=2*acceleration*distance
solve for distance.
I am stay a student
To find the distance the boy would reach a speed of 8.00 m/s, we can use the equations of motion.
The equation to calculate the final velocity, given the initial velocity and acceleration, is:
v = u + at
Where:
v = final velocity (8.00 m/s)
u = initial velocity (0 m/s, since the boy started from rest)
a = acceleration (1.30 m/s^2)
t = time taken to reach the final velocity
To find 't', we need to rearrange the equation and solve for 't':
t = (v - u) / a
Substituting the given values:
t = (8.00 m/s - 0 m/s) / 1.30 m/s^2
t = 6.15 seconds
Now, we can use another equation of motion to find the distance traveled:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity (0 m/s)
t = time taken (6.15 s)
a = acceleration (1.30 m/s^2)
Substituting the values:
s = (0 m/s)(6.15 s) + (1/2)(1.30 m/s^2)(6.15 s)^2
s = 0 + 0.50(1.30 m/s^2)(37.8225 s^2)
s = 0.50(1.30 m/s^2)(37.8225 s^2)
s = 0.50(48.91025 m)
s = 24.45512 (approximately 24.5 m)
Therefore, the boy would reach a speed of 8.00 m/s in approximately 24.5 meters.