An airplane lands and starts down the runway at a southwest velocity of 43 m/s. What constant acceleration allows it to come to a stop in 1.2 km?Answer is in ___ m/s2.
2 answers
Vf^2=Vi^2+2a*distance
V = sqrt (2 a X)
X = 1200 m
V = 43 m/s
Solve for a. Add a minus sign because there must be deceleration. Or make the direction of the acceleration 180 degrees opposite from the landing direction.
X = 1200 m
V = 43 m/s
Solve for a. Add a minus sign because there must be deceleration. Or make the direction of the acceleration 180 degrees opposite from the landing direction.
0 answers