The rocket-driven sled sonic wind no. 2, used for investigating the physiological effects of large accelerations, runs on a straight, level track that is 1080 m long. starting from rest, it can reach a speed of 1560 km/h in 1.70 s.
a) compute the acceleration in m/s^2.
b) compute the acceleration in gís.
c) what is the distance covered in 1.70s?

A) acceleration = [change in velocity] / [time interval] = [final velocity - initial velocity] / t = 

acceleration = [1560 km/h -0] / 1.7 s = 1560 km/h / 1.7 s

convert the velocity to m/s: 1560 km/h * 1000 m/ km * 1h/3600 s = 433.33 m/s

acceleration = 433.33 m/s / 1.7 s = 254.9 m/s^2

b) Acceleration in terms of g

 acceleration = 254.9 m/s^2 /[ 9.8 m/s^2 / g] = 26 g.

This is the acceleration is 26 times the gravitational acceleration.

c) distance

d = a*t^2 / 2 = 254.9 m/s^2 * (1.70s)^2 / 2 = 88.2 m

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