for acceleration we can define that rate of change in velocity is know as acceleration
So whenever velocity of train is changing with time we can say train is accelerating
Now here if initially train is standstill then after some time its speed is 5 m/s
so here the train is accelerated first time
Then on straight path its speed changed from 5 m/s to 10 m/s so here train gets accelerated second time
After this train chugged around a curve with same speed 10 m/s
SO here since train is moving in curve so here its direction of velocity is continuously changing and this type of acceleration is known as centripetal acceleration
SO this is accelerated Third time
Then its speed decreases and it comes to speed of 5 m/s from 10 m/s
So here it is acceleration of train for Fourth time
Then finally train comes to stop so again its speed changed from 5 m/s to 0
so this is acceleration of train Fifth time
So total train will accelerate 5 times in whole path
Answer:
Making a Hypothesis
Explanation:
-Research the subject of your question. Review the literature and find out as much as you can about previous information and discoveries surrounding your question.
-Develop an educated guess that answers your initial question. This is your hypothesis. Make a prediction based on your hypothesis and state it as a cause-effect relationship.
To solve the problem we will first start considering the Pressure given the hydrostatic definition of the product between the density, the gravity and the depth. We will define the area where the liquid acts and later we will use the definition of the force as a product between the pressure and the area to calculate the force given in the two depths. The gauge pressure at the depth x will be
This pressure acts on the strip of area
The force acting on that strip is given by,
To evaluate the force, we will then consider the integral of the pressure as a function of the Area, or the integral of the previously found terms.
Evaluating at the initial depth of 1.8m and the final depth of 4.4 we have then that,
Therefore the Net force will be
