Sanjay's air boat starts from rest and accelerates at 9.4 m/s2 for a distance of 245 meters. How fast is it moving at the end of the 245 meters?
To determine the final velocity of Sanjay's air boat at the end of the given distance, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
- v = final velocity (what we need to find)
- u = initial velocity (starts from rest, so it is 0 m/s)
- a = acceleration (given as 9.4 m/s^2)
- s = distance (given as 245 meters)
Substituting the known values into the equation, we have:
v^2 = 0^2 + 2 * 9.4 * 245
Simplifying,
v^2 = 0 + 2 * 9.4 * 245
v^2 = 0 + 2 * 2303
v^2 = 4606
Taking the square root of both sides to solve for v:
v = √4606
v ≈ 67.85 m/s
Therefore, the air boat is moving at approximately 67.85 m/s at the end of the 245 meters.
245 = 1/2 * 9.4 * t^2
solve for t (time)
velocity = a * t = 9.4 * t ... m/s