A particle moves along a straight line between two points P and Q with constant acceleration 0.8 m s–2. Its velocity at Q is 1.2 m s–1 greater than the velocity at P. If the distance PQ is 48 m, find the velocity at P. How long after passing P does it take the velocity to reach 48 m s–1.

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j