Sanjay's air boat starts from rest and accelerates at 8.7 m/s2 for a distance of 176 meters. How fast is it moving at the end of the 176 meters?
d = (1/2) a t^2
176 = 4.35 t^2 solve for t
v = a t = 8.7 t
To find the final speed of Sanjay's air boat, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity (which is 0 in this case since the air boat starts from rest)
a = acceleration
s = distance
Given that the air boat starts from rest (u = 0), accelerates at 8.7 m/s^2 (a = 8.7 m/s^2), and travels a distance of 176 meters (s = 176 m), we can substitute these values into the equation:
v^2 = 0^2 + 2 * 8.7 * 176
Simplifying the equation:
v^2 = 2 * 8.7 * 176
v^2 = 3057.6
To find the final velocity (v), we calculate the square root of both sides of the equation:
v = √3057.6
Using a calculator, we find that v is approximately 55.37 m/s.
Therefore, the air boat is moving at approximately 55.37 m/s at the end of the 176 meters.