You should have found that at the end of the second from t = 0 to t = 1.0 s the probe was moving at (3 m/s, down) and that it has gone down 1,5 m from where it started. Suppose
that in the next second (t = 1.0 s to t = 2.0 s) its retro thrusters are still not firing. So, during this second it is falling with an acceleration of (3 m/s2, down) again. Match the
following quantities for the probe to their correct values. In all cases we are considering "initial" to mean the beginning of this second (t = 1.0 s) and "final" to mean the end of this
second (t = 2.0 s).
Initial velocity?
Final velocity?
Average velocity?
Displacement during this second?
Displacement since it was released?
Answers
Answer:
Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its size, its velocity, and the fluid it is in. For most large objects such as cyclists, cars, and baseballs not moving too slowly, the magnitude of the drag force FDFD is proportional to the square of the speed of the object. We can write this relationship mathematically as FD∝v2.FD∝v2. When taking into account other factors, this relationship becomes
FD=12CρAv2,FD=12CρAv2,
where C is the drag coefficient, A is the area of the object facing the fluid, and ρρ is the density of the fluid. (Recall that density is mass per unit volume.) This equation can also be written in a more generalized fashion as FD=bv2,FD=bv2, where b is a constant equivalent to 0.5CρA.0.5CρA. We have set the exponent n for these equations as 2 because when an object is moving at high velocity through air, the magnitude of the drag force is proportional to the square of the speed. As we shall see in Fluid Mechanics, for small particles moving at low speeds in a fluid, the exponent n is equal to 1.