Sanjay's air boat starts from rest and accelerates at 4.6 m/s2 for a distance of 219 meters. How fast is it moving at the end of the 219 meters?
d = Vo t + (1/2) a t^2
219 = 0 + 2.3 t^2
solve for t
v = Vo + a t
v = 0 + 4.6 t
To find the final velocity of Sanjay's airboat at the end of the 219 meters, we can use the equation of motion:
v^2 = u^2 + 2as
where
v = final velocity (what we want to find)
u = initial velocity (which is 0 m/s since the airboat starts from rest)
a = acceleration (4.6 m/s^2)
s = distance (219 meters)
Now, we can plug in the given values into the equation and find the final velocity:
v^2 = 0^2 + 2 * 4.6 * 219
Simplifying the equation:
v^2 = 0 + 1008.4
Taking the square root of both sides to solve for v:
v = √1008.4
Calculating the value:
v ≈ 31.75 m/s
Therefore, the airboat is moving at approximately 31.75 m/s at the end of the 219 meters.