When you talk about rate, you will expect that it will be in terms of a time unit. It measures how fast it is going. So, you would expect that the denominator is in time units. For the movement, you can measure this with either distance or velocity.
So, for the first variety, you would need distance and time to measure the rate of how far you go at a certain time. It is also called as velocity. For the second variety, you would need velocity and time to measure the rate of how fast you are going at a certain interval. It is also called as acceleration.
Answer:
Potential energy =mass* acceleration due to gravity * height
mass*acceleration due to gravity =weight
hence potential energy of the puppy= weight * height
=18*2
=36 joule
The heat released by the water when it cools down by a temperature difference

is

where
m=432 g is the mass of the water

is the specific heat capacity of water

is the decrease of temperature of the water
Plugging the numbers into the equation, we find

and this is the amount of heat released by the water.
To determine the height of the object given the time, we simply use the given relation between height and time in the problem statement. It is given as:
h = -16t^2 + 127t
We substitute 55 seconds to t and obtain,
h = -16(55)^2 + 127(55)
h = - 41415
Objects will accelerate more by 10 (m/s)2