Question
A jet lands at 60.2 m/s, the pilot applying the brakes 2.28 s after landing. Find the acceleration needed to stop the jet within 5.83 102 m after touchdown.
Acceleration answered in (m/s2)
Acceleration answered in (m/s2)
Answers
oobleck
If it takes t seconds to stop after braking, then
60.2 + at = 0
t = -60.2/a
Now, using the distance,
60.2*2.28 - 60.2t + 1/2 a t^2 = 583
60.2*2.28 - 60.2^2/a + 1/2 * 60.2^2/a = 583
a = -4.06 m/s^2
60.2 + at = 0
t = -60.2/a
Now, using the distance,
60.2*2.28 - 60.2t + 1/2 a t^2 = 583
60.2*2.28 - 60.2^2/a + 1/2 * 60.2^2/a = 583
a = -4.06 m/s^2