The discussion centers on a physics problem involving a helicopter lifting off with a constant acceleration of 5.0 m/s² for 11.0 seconds. The correct formula to calculate the total distance traveled is y = y₀ + v₀yt + 1/2*a*t². The initial calculations yielded a distance of 300 meters, while the accurate answer is 460 meters, highlighting the importance of precise arithmetic in physics calculations.
PREREQUISITESStudents studying physics, educators teaching kinematics, and anyone interested in mastering calculations related to motion and acceleration.
student34 said:Homework Statement
If you are in a helicopter and the helicopter lifts off at a constant acceleration of 5.0m/s^2 for 11.0 seconds, what is the total distance traveled from the ground?
Homework Equations
y = yo + voyt + 1/2*ax*t^2
The Attempt at a Solution
y = 0 + 0 + 1/2(5m/s^2)(11^2)
y = 3*10^2m
But, apparently the actual answer is 4.6*10^2m.
I'll check mine instead !Simon Bridge said:Guess I have to check mine too :(