Can anyone help me?please tell me how to do this problem?

Question

landing with a speed of 115m/s, and traveling due south, a jet comes to rest in 700m. assuming the jet slows with constant acceleration, find the magnitude and direction of its acceleration.

Answer 1

Use the constant acceleration equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, s is the distance and a is the acceleration. For these purposes say that south is positive (this problem is one-dimensional).

Plug in the numbers; v=0, u=115, s=700 (all given, and we are working in SI units so no conversion factors) leads to 0=115^2 + 2 x a x 700. Simplify and collect terms; 1400a = -115^2 or a = -(115^2)/1400 (a is negative, meaning that it is in the opposite direction to the motion). This works out to 9.446 m/s^2, northwards.

Answer 2

Direction of the acceleration would be due North in opposition to the direction of motion.    Speed changes from 115 m/s to 0 m/s over an interval of 700 m.  We can

assume average speed is therefore 115/2 m/s and

700 m/(115m/2s) = 12.1739 s to complete stop.  

Since speed is reduced from 115 m/s to 0 m/s over that

period, divide the 115 m/s by the 12.1739 s to determine

the magnitude  , that's  9.446 m/s²  with direction due North