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 x 102 m after touchdown. Answer in m/s2.

The initial velocity of the jet is 60.2 m/s. The final velocity is 0 m/s, since the jet needs to come to a stop. The distance it travels during braking is 5.83 x 10^2 m. The time taken to stop, t = 2.28 s. Using the formula, v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time, 0 = 60.2 + a x 2.28 a = -26.4 m/s^2 (Note the negative sign indicates that the acceleration is in the opposite direction of the motion of the jet.) To find the acceleration needed to stop the jet within 5.83 x 10^2 m, we use the formula, v^2 = u^2 + 2as, 0 = (60.2)^2 + 2 x a x 5.83 x 10^2 a = -7.74 m/s^2 Again, the negative sign indicates that the acceleration is opposite to the motion of the jet. Therefore, the acceleration needed to stop the jet within 5.83 x 10^2 m after touchdown is 7.74 m/s^2.