Recall that

where
and
are the initial and final velocities, respecitvely;
is the acceleration; and
is the change in position.
So we have


(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)
Answer:
Surely Achilles will catch the Tortoise, in 400 seconds
Explanation:
The problem itself reduces the interval of time many times, almost reaching zero. However, if we assume the interval constant, then it is clear that in two hours Achilles already has surpassed the Tortoise (20 miles while the Tortoise only 3).
To calculate the time, we use kinematic expression for constant speed:

The moment that Achilles catch the tortoise is found by setting the same final position for both (and same time as well, since both start at the same time):

Answer:
Explanation:
Potential energy on the surface of the earth
= - GMm/ R
Potential at height h
= - GMm/ (R+h)
Potential difference
= GMm/ R - GMm/ (R+h)
= GMm ( 1/R - 1/ R+h )
= GMmh / R (R +h)
This will be the energy needed to launch an object from the surface of Earth to a height h above the surface.
Extra energy is needed to get the same object into orbit at height h
= Kinetic energy of the orbiting object at height h
= 1/2 x potential energy at height h
= 1/2 x GMm / ( R + h)
Answer:
The soda is being sucket out at a rate of 3.14 cubic inches/second.
Explanation:
R= 2in
S= π*R²= 12.56 inch²
rate= 0.25 in/sec
rate of soda sucked out= rate* S
rate of soda sucked out= 3.14 inch³/sec
Answer:
130 km at 35.38 degrees north of east
Explanation:
Suppose the HQ is at the origin (x = 0, y = 0)
So the coordinates of the helicopter after the 1st flight is


After the 2nd flight its coordinate would be:


So in order to fly back to its HQ it must fly a distance and direction of
north of east