A diver having mass m climbs up the diving board.
We know that Gravitational potential energy is given as <span>P<span>EG</span>=mgΔh</span>
What changes is his Gravitational potential energy due to change of height <span>Δh</span> with reference to the ground/water level.
While standing on the diving board his velocity is zero. As such kinetic energy is also zero.
Once he jumps off the springboard we see he gets additional energy from the springboard and falls down under action of gravity g. Due to decrease of height above the ground level Gravitational potential energy decreases and gets converted in to his kinetic energy. <span>1/2m<span>v2</span></span>.
While in air he encounters air resistance. Some of his energy is spent in overcoming this resistance. Gets converted in to kinetic and thermal energy of surrounding air and his body.
Once diver reaches the water, we see water splashing and hear noise of splash. Thereafter the diver comes to rest. Now his potential energy becomes zero. And converted kinetic energy has been converted in to kinetic energy, heat energy and sound energy of water.
As such energy transformation equation looks like
<span><span>Gravitational PE+Elastic PE of springboard</span><span>→Kinetic energy of air and water+Sound energy of splash+thermal energy</span></span>
We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.
Answer:
Part a)
Part b)
Explanation:
Part a)
Constant speed by which the student will run is given as
now after some time if student is going to overtake the position of bus
so here the final positions will be same
so we have
so it is
So student will run the total distance
Part b)
Speed of bus when student reach the bus is given as
